Unit 1 Kinematics 1.m Projectile Motion Answer Key

Unit 1 kinematics 1.mprojectile motion answer key is a vital resource for students who are beginning to explore the physics of objects moving under the influence of gravity alone. This guide breaks down the fundamental concepts, presents the essential equations, walks through a detailed solution process, and provides a set of practice problems with complete answers so learners can check their work and build confidence. By mastering the material covered here, you will be able to predict the trajectory, range, maximum height, and time of flight of any projectile launched near Earth’s surface, assuming air resistance is negligible.

Understanding Projectile Motion

Projectile motion occurs when an object is launched into the air and moves along a curved path under the sole influence of gravity. The motion can be analyzed by separating it into two independent components: horizontal (x‑direction) and vertical (y‑direction). Because there is no horizontal acceleration (assuming no air resistance), the horizontal velocity remains constant. Vertically, the object experiences a constant acceleration downward equal to g ≈ 9.8 m/s². This independence allows us to treat each component with simple kinematic equations and then recombine the results to describe the overall trajectory.

Key characteristics of projectile motion include:

• The trajectory is a parabola when plotted in the x‑y plane.
• The time of flight depends only on the vertical motion.
• The horizontal range is maximized when the launch angle is 45° (for equal launch and landing heights).
• The maximum height is reached when the vertical velocity component becomes zero.

Essential Equations

To solve any projectile problem, start with the following kinematic formulas, applying them separately to the horizontal and vertical axes:

Horizontal motion (constant velocity):
[ x = v_{0x},t ]
where (v_{0x}=v_0\cos\theta) is the initial horizontal speed.

Vertical motion (constant acceleration): [ y = v_{0y},t - \frac{1}{2}gt^{2} ]
[ v_{y}=v_{0y}-gt ]
[ v_{y}^{2}=v_{0y}^{2}-2gy ]
with (v_{0y}=v_0\sin\theta) as the initial vertical speed.

From these, we derive the most useful projectile‑specific expressions:

• Time of flight (when launch and landing heights are equal):
  [ T = \frac{2v_0\sin\theta}{g} ]

• Maximum height:
  [ H = \frac{(v_0\sin\theta)^{2}}{2g} ]

• Horizontal range:
  [ R = \frac{v_0^{2}\sin(2\theta)}{g} ]

Remember that these formulas assume the projectile starts and ends at the same vertical level. If the launch and landing heights differ, you must use the full vertical displacement equation and solve for time using the quadratic formula.

Step‑by‑Step Solution Guide

Follow this systematic approach when tackling a projectile motion problem:

1. Read the problem carefully and identify what is given (initial speed, launch angle, height, etc.) and what is asked for (range, height, time, velocity at a point, etc.).
2. Draw a diagram showing the launch point, trajectory, and coordinate axes. Label known quantities and the unknown you need to find.
3. Resolve the initial velocity into its components:
   (v_{0x}=v_0\cos\theta) and (v_{0y}=v_0\sin\theta).
4. Choose the appropriate equations for horizontal and vertical motion based on the known and unknown variables.
5. Solve for time first if it is not directly given, using the vertical motion equation (often a quadratic).
6. Plug the time into the horizontal equation to find range or horizontal displacement, or into the vertical equation to find height or final vertical velocity.
7. Check units and ensure the answer is physically reasonable (e.g., time positive, range not negative unless the problem specifies a backward launch).
8. If needed, compute the final velocity vector by combining the final horizontal and vertical components:
   (v_f = \sqrt{v_{fx}^{2}+v_{fy}^{2}}) and direction (\phi = \tan^{-1}(v_{fy}/v_{fx})).

Applying this method consistently reduces errors and builds a clear mental model of the projectile’s behavior.

Common Mistakes to Avoid

Even experienced students can slip up on projectile problems. Watch out for these frequent pitfalls:

• Mixing up sine and cosine when resolving the initial velocity. Remember: the component adjacent to the angle uses cosine, the opposite uses sine.
• Forgetting that horizontal acceleration is zero and mistakenly applying (a_x = g) in the x‑direction.
• Using degrees in calculators set to radians (or vice versa) when evaluating trigonometric functions.
• Neglecting the sign of gravity; always treat (g) as positive magnitude and insert a minus sign in the vertical equations to indicate downward acceleration.
• Assuming the time of flight formula works for unequal launch and landing heights; it does not—use the full quadratic solution instead.
• Overlooking air resistance when the problem explicitly states to ignore it; if air resistance is mentioned, the simple parabolic model no longer applies.

Being aware of these issues will help you catch mistakes before they affect your final answer.

Practice Problems with Answer Key

Below are five representative problems that cover the core concepts of unit 1 kinematics 1.m projectile motion. Attempt each on your own, then compare your solution to the provided answer key.

Problem 1

A soccer ball is kicked from ground level with an initial speed of 20 m/s at an angle of 30° above the horizontal. (a) Calculate the time of flight. (b) Determine the horizontal range.
(c) Find the maximum height reached.

Problem 2

A projectile is launched from the top of a 45 m tall cliff with an initial velocity of 25 m/s at an angle of 50° above the horizontal.
(a) How long does it take to hit the ground below the cliff?
(b) What is the horizontal distance from the base of the cliff to the impact point?
(c) What is the magnitude and direction of the velocity just before impact?

Problem 3

An archer shoots an arrow horizontally from a height of 1.5 m with a speed of 12 m/s.
(a) How long does the arrow stay in the air before hitting the ground?
(b) How far horizontally does it travel before impact?

Problem 4

Problem 4

A cannon fires a projectile from ground level with an initial speed of 40 m/s at an angle of 60° above the horizontal.
(a) Compute the time of flight.
(b) Determine the horizontal range.
(c) Find the maximum height attained.
(d) What are the horizontal and vertical components of the velocity at the instant the projectile is halfway to its maximum height?

Problem 5

A baseball is thrown from a height of 2.0 m above the ground with an initial speed of 18 m/s directed 25° below the horizontal (i.e., the throw is downward).
(a) How long does it take the ball to reach the ground?
(b) What is the horizontal distance traveled before impact?
(c) What is the speed and direction of the velocity just before it strikes the ground?

Answer Key

Problem 1

• Given: (v_0 = 20\text{ m/s},\ \theta = 30^\circ)
• (v_{0x}=v_0\cos\theta = 20\cos30^\circ = 20(0.8660)=17.32\text{ m/s})
• (v_{0y}=v_0\sin\theta = 20\sin30^\circ = 20(0.5)=10.0\text{ m/s})

(a) Time of flight (launch and landing at same height):
(t_{flight}= \frac{2v_{0y}}{g}= \frac{2(10.0)}{9.81}=2.04\text{ s})

(b) Horizontal range:
(R = v_{0x},t_{flight}=17.32\times2.04 = 35.3\text{ m})

(c) Maximum height:
(H_{max}= \frac{v_{0y}^2}{2g}= \frac{10.0^2}{2(9.81)} = 5.10\text{ m})

Problem 2

• Given: (v_0 = 25\text{ m/s},\ \theta = 50^\circ,\ h_0 = 45\text{ m}) (launch above ground)
• (v_{0x}=25\cos50^\circ = 25(0.6428)=16.07\text{ m/s}) - (v_{0y}=25\sin50^\circ = 25(0.7660)=19.15\text{ m/s})

(a) Time to hit ground: Solve (y(t)=h_0+v_{0y}t-\frac12gt^2=0).
(0 = 45 + 19.15t - 4.905t^2) → (4.905t^2 -19.15t -45 =0)
Using quadratic formula:
(t = \frac{19.15 + \sqrt{19.15^2+4(4.905)(45)}}{2(4.905)}) (positive root)
(t = \frac{19.15 + \sqrt{366.7+882.9}}{9.81}= \frac{19.15 + \sqrt{1249.6}}{9.81})
(\sqrt{1249.6}=35.35) → (t = \frac{19.15+35.35}{9.81}= \frac{54.5}{9.81}=5.56\text{ s})

(b) Horizontal distance:
(x = v_{0x}t = 16.07\times5.56 = 89.3\text{ m})

(c) Velocity just before impact:
(v_x = v_{0x}=16.07\text{ m/s}) (unchanged)
(v_y = v_{0y} - gt = 19.15 - 9.81(5.56)=19.15-54.5 = -35.35\text{ m/s}) (downward)

Magnitude: (v = \sqrt{v_x^2+v_y^2}= \sqrt{16.07^2+(-35.35)^2}= \sqrt{258.2+1249.6}= \sqrt{1507.8}=38.8\text{ m/s})

Answer Key

Problem 1

• Given: (v_0 = 20\text{ m/s},\ \theta = 30^\circ)
• (v_{0x}=v_0\cos\theta = 20\cos30^\circ = 20(0.8660)=17.32\text{ m/s})
• (v_{0y}=v_0\sin\theta = 20\sin30^\circ = 20(0.5)=10.0\text{ m/s})

(a) Time of flight (launch and landing at same height): (t_{flight}= \frac{2v_{0y}}{g}= \frac{2(10.0)}{9.81}=2.04\text{ s})

(b) Horizontal range: (R = v_{0x},t_{flight}=17.32\times2.04 = 35.3\text{ m})

(c) Maximum height: (H_{max}= \frac{v_{0y}^2}{2g}= \frac{10.0^2}{2(9.81)} = 5.10\text{ m})

Problem 2

• Given: (v_0 = 25\text{ m/s},\ \theta = 50^\circ,\ h_0 = 45\text{ m}) (launch above ground)
• (v_{0x}=25\cos50^\circ = 25(0.6428)=16.07\text{ m/s})
• (v_{0y}=25\sin50^\circ = 25(0.7660)=19.15\text{ m/s})

(a) Time to hit ground: Solve (y(t)=h_0+v_{0y}t-\frac12gt^2=0). (0 = 45 + 19.15t - 4.905t^2) → (4.905t^2 -19.15t -45 =0) Using quadratic formula: (t = \frac{19.15 + \sqrt{19.15^2+4(4.905)(45)}}{2(4.905)}) (positive root) (t = \frac{19.15 + \sqrt{366.7+882.9}}{9.81}= \frac{19.15 + \sqrt{1249.6}}{9.81}) (\sqrt{1249.6}=35.35) → (t = \frac{19.15+35.35}{9.81}= \frac{54.5}{9.81}=5.56\text{ s})

(b) Horizontal distance: (x = v_{0x}t = 16.07\times5.56 = 89.3\text{ m})

(c) Velocity just before impact: (v_x = v_{0x}=16.07\text{ m/s}) (unchanged) (v_y = v_{0y} - gt = 19.15 - 9.81(5.56)=19.15-54.5 = -35.35\text{ m/s}) (downward) Magnitude: (v = \sqrt{v_x^2+v_y^2}= \sqrt{16.07^2+(-35.35)^2}= \sqrt{258.2+1249.6}= \sqrt{1507.8}=38.8\text{ m/s})

Problem 3

(a) Does the arrow stay in the air before hitting the ground? Yes, the arrow remains in the air for a duration determined by its initial velocity and the acceleration due to gravity. The time of flight is calculated as the total time the arrow is in the air, which is equal to the time it takes to reach its maximum height and return to the same height.

(b) How far horizontally does it travel before impact? The horizontal range is calculated by multiplying the initial horizontal velocity by the time of flight. Since there is no horizontal acceleration (ignoring air resistance), the horizontal velocity remains constant throughout the flight. Therefore, the horizontal distance traveled is equal to the horizontal component of the initial velocity multiplied by the time of flight.

Problem 4

Problem 4

A cannon fires a projectile from ground level with an initial speed of 40 m/s at an angle of 60° above the horizontal. (a) Compute the time of flight. (b) Determine the horizontal range. (c) Find the maximum height attained. (d) What are the horizontal and vertical components of the velocity

Problem 4 (continued)

(a) Flight duration
The projectile leaves the barrel with a speed of 40 m s⁻¹ at 60° above the horizontal.
The vertical component of the launch speed is

[ v_{0y}=40\sin 60^{\circ}=40(0.8660)=34.64\ \text{m s}^{-1}. ]

Because launch and landing heights coincide, the total time aloft follows from the symmetric nature of the trajectory:

[T=\frac{2v_{0y}}{g}= \frac{2(34.64)}{9.81}=7.07\ \text{s}. ]

(b) Horizontal span
The horizontal component of the initial velocity is

[ v_{0x}=40\cos 60^{\circ}=40(0.5000)=20.0\ \text{m s}^{-1}, ]

which stays constant throughout the flight (air‑drag neglected). Multiplying this steady speed by the total time gives the range:

[ R = v_{0x},T = 20.0 \times 7.07 = 1.41\times10^{2}\ \text{m}. ]

(c) Peak altitude
At the apex the vertical speed momentarily drops to zero. Using

[ H_{\max}= \frac{v_{0

y}^{2}}{2g} ]

yields

[ H_{\max}= \frac{(34.64)^{2}}{2(9.81)} = \frac{1200.0}{19.62} = 61.2\ \text{m}. ]

(d) Velocity components at impact
The horizontal component remains unchanged: (v_x = 20.0\ \text{m s}^{-1}).
The vertical component at landing is equal in magnitude but opposite in direction to the initial vertical component:

[ v_y = -v_{0y} = -34.64\ \text{m s}^{-1}. ]

The speed just before striking the ground is

[ v = \sqrt{v_x^{2} + v_y^{2}} = \sqrt{20.0^{2} + (-34.64)^{2}} = \sqrt{400 + 1200} = \sqrt{1600} = 40.0\ \text{m s}^{-1}. ]

Thus, the projectile lands with the same speed it had at launch, but directed downward at an angle of (60^{\circ}) below the horizontal.