27 Projectile motion – problems and solutions

1. A bullet fired at an angle θ = 60^{o }with a velocity of 20 m/s. Acceleration due to gravity is 10 m/s^{2}. What is the time interval to reach the maximum height?

__Known :__

The initial velocity of bullet (v_{o}) = 20 m/s

Angle (θ) = 60^{o}C

Acceleration due to gravity (g) = 10 m s^{–2}

__Wanted :__ The time interval to reach the maximum height

__Solution :__

The initial velocity at the horizontal direction (x axis) :

v_{ox} = v_{o} cos 60^{o} = (20)(0.5) = 10 m/s

The initial velocity at the vertical direction (y axis) :

v_{oy} = vo sin 60^{o} = (20)(0.5√3) = 10√3 m/s

The time interval to reach the maximum height, calculated using this equation :

v_{ty} = v_{oy} + g t

v_{ty} = the final velocity in the vertical direction = the final velocity at the highest point = 0 m/s

v_{oy }= the initial velocity at the horizontal direction = 10√3 m/s

g = acceleration due to gravity = 10 m/s^{2}

t = time interval

The time interval :

v_{ty} = v_{oy} + g t

0 = 10√3 – 10 t

10√3 = 10 t

t = 10√3 / 10

t = √3 seconds

2. An object projected at an angle. The height of the object is the same when the time interval = 1 second and 3 seconds. What is the time interval the object in air.

__Solution :__

The object in the air for 4 seconds.

3. An aircraft is moving horizontally with a speed of 50 m/s. At the height of 2 km, an object is dropped from the aircraft. Acceleration due to gravity = 10 m/s2, what is the time interval before the object hits the ground.

__Known :__

Height = 2 km = 2000 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ The time interval (t)

__Solution :__

h = 1/2 g t^{2}

2000 = 1/2 (10) t^{2}

2000 = 5 t^{2}

t^{2 }= 2000/5 = 400

t = √400 = 20 seconds

4. A kicked football leaves the ground at an angle θ = 45^{o }with the horizontal has an initial speed of 25 m/s. Determine the distance of X. Acceleration due to gravity is 10 m/s^{2}.

__Known :__

Initial speed (v_{o}) = 25 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

Angle (θ) = 45^{o}

__Wanted :__ X

__Solution :__

The horizontal component of the initial velocity :

v_{ox} = v_{o} cos θ = (25 m/s)(cos 45^{o}) = (25 m/s)(0.5√2) = 12.5√2 m/s

The vertical component of the initial velocity :

v_{oy }= v_{o} sin θ = (25 m/s)(sin 45^{o}) = (25 m/s)(0.5√2) = 12.5√2 m/s

Projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at constant velocity and the y motion occurs at constant acceleration of gravity.

**Time in the air**** ****(t) :**

The time in air calculated with the equation of the upward vertical motion.

Choose upward direction as positive and downward direction as negative.

__Known :__

The initial velocity (v_{o}) = 12.5√2 m/s (upward direction, positive)

Acceleration due to gravity (g) = -10 m/s^{2 }(downward direction, negative)

Height (h) = 0

__Wanted :__ Time interval (t)

__Solution :__

h = v_{o} t + 1/2 g t^{2}

0 = (12.5√2) t + 1/2 (-10) t^{2}

0 = 12.5√2 t – 5 t^{2}

12.5√2 t = 5 t^{2}

12.5√2 = 5 t

t = 12.5√2 / 5

t = 2.5√2 seconds

**The horizontal distance ****(X) :**

Calculated using the equation of the uniform linear motion with constant velocity.

__Known :__

Velocity (v) = 12.5√2 m/s

Time interval (t) = 2.5√2 seconds

__Wanted :__ Distance

__Solution :__

d = v t = (12.5√2)(2.5√2) = (12.5)(2.5)(2) = 62.5 meters

5. An object projected upward at an angle θ = 30^{o }with the horizontal has an initial speed of 20 m/s. Acceleration due to gravity is 10 m/s^{2}. Determine the maximum height.

__Known :__

The initial velocity (v_{o}) = 20 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

Angle (θ) = 30^{o}

__Wanted ____:__ The maximum height

__Solution :__

First, find the vertical component of the initial velocity (v_{oy}) :

v_{oy} = v_{o} sin 30^{o }= (20)(sin 30^{o}) = (20)(0.5) = 10 m/s

Calculate the maximum height. Choose upward direction as positive and downward direction as negative.

__Known :__

Acceleration due to gravity (g) = -10 m/s^{2 }*(**downward **direction, negative**)*

The vertical component of the initial velocity (v_{oy}) = 10 m/s *(**upward direction, positive**)*

Velocity at the maximum height (v_{ty}) = 0

__Wanted :__ The maximum height (h)

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 g h

0^{2} = 10^{2} + 2 (-10) h

0 = 100 – 20 h

100 = 20 h

h = 100/20

h = 5 meters

The maximum height is 5 meters.

6. An object is thrown at a certain elevation angle. The height of the object same after 1 second and 3 seconds. Determine time in air.

A. 3.6 s

B. 4.0 s

C. 5.6 s

D. 6.4 s

Solution

Time in air = 4 seconds.

The correct answer is B.

7. An aircraft is moving horizontally with the speed of 50 m/s. When the aircraft at the height of 2 km, an object free fall from the aircraft. Determine the type of the motion.

A. Free fall motion

B. Floating motion

C. Horizontal motion

D. Projectile motion

Solution :

The object is dropped from the moving plane because it has the same speed as the plane’s speed, that is 50 m/s. Movement of objects is not like free fall motion but parabolic motion. The case is the same as you are dropping objects from inside a moving car.

The correct answer is D.

8. A ball is thrown horizontally at 15 m/s from a cliff 60 meters high. How long does it take to hit the ground?

Solution: Using \( h = \frac{1}{2} g t^2 \), we find the time is \( t = \sqrt{\frac{2h}{g}} \approx 3.5\ \text{s} \).

9. A projectile is fired at an angle of 30° above the horizontal with an initial speed of 20 m/s. What is the maximum height reached?

Solution: Using \( h = \frac{v^2 \sin^2 \theta}{2g} \), the maximum height is \( h \approx 10.2\ \text{m} \).

10. A stone is thrown horizontally at 10 m/s from a tower 80 meters tall. Find the horizontal distance it travels before hitting the ground.

Solution: Using the time found in a similar way to Problem 1, the horizontal distance is \( d = vt \approx 40\ \text{m} \).

11. A cannonball is fired at 40 m/s at an angle of 45°. Find the time of flight.

Solution: Using \( t = \frac{2v \sin \theta}{g} \), the time of flight is \( t \approx 5.8\ \text{s} \).

12. A baseball is thrown at an angle of 60° with a velocity of 12 m/s. Find the horizontal range.

Solution: Using \( R = \frac{v^2 \sin 2\theta}{g} \), the range is \( R \approx 14.0\ \text{m} \).

13. A projectile is launched with an initial velocity of 50 m/s at 37° above the horizontal. What is the vertical velocity component?

Solution: The vertical component is \( v_y = v \sin \theta \approx 30\ \text{m/s} \).

14. A projectile is launched horizontally at 20 m/s from a height of 100 meters. What is the vertical velocity just before it hits the ground?

Solution: Using \( v_y = \sqrt{2gh} \), the vertical velocity is \( v_y \approx 44.7\ \text{m/s} \).

15. A rock is thrown at an angle of 25° above the horizontal with an initial speed of 15 m/s. What are the horizontal and vertical components of the velocity?

Solution: The horizontal component is \( v_x = v \cos \theta \approx 13.4\ \text{m/s} \), and the vertical component is \( v_y \approx 6.4\ \text{m/s} \).

16. A soccer ball is kicked with an initial speed of 30 m/s at an angle of 40° above the horizontal. What is its horizontal velocity component?

Solution: The horizontal component is \( v_x = v \cos \theta \approx 22.9\ \text{m/s} \).

17. A golf ball is hit with an initial speed of 70 m/s at an angle of 20°. What is the time of flight?

Solution: Using the time of flight equation, the time is \( t \approx 4.9\ \text{s} \).

18. A projectile is fired from the ground with a speed of 25 m/s at 53° above the horizontal. What is its initial vertical velocity component?

Solution: The vertical component is \( v_y = v \sin \theta \approx 20\ \text{m/s} \).

19. A baseball is thrown with an initial speed of 20 m/s at an angle of 50°. What is the maximum height?

Solution: Using the equation for maximum height, the height is \( h \approx 15.3\ \text{m} \).

20. A bullet is fired horizontally with a velocity of 200 m/s from a height of 10 meters. How long does it take to hit the ground?

Solution: Using the time equation, the time is \( t \approx 1.4\ \text{s} \).

21. A cannonball is fired at 45 m/s at an angle of 30°. Find the range.

Solution: Using the range equation, the range is \( R \approx 88.2\ \text{m} \).

22. A basketball is thrown at an angle of 75° with a velocity of 10 m/s. Find the horizontal range.

Solution: Using the range equation, the range is \( R \approx 5.3\ \text{m} \).

23. A projectile is launched with an initial velocity of 30 m/s at 22° above the horizontal. What is the vertical velocity just before it hits the ground?

Solution: Using the vertical velocity equation, the vertical velocity is \( v_y \approx 11.4\ \text{m/s} \).

24. A rock is thrown horizontally at 8 m/s from a tower 40 meters tall. What is the horizontal distance it travels?

Solution: Using the horizontal distance equation, the distance is \( d \approx 16\ \text{m} \).

25. A soccer ball is kicked with an initial speed of 12 m/s at an angle of 30° above the horizontal. What are the horizontal and vertical components of the velocity?

Solution: The horizontal component is \( v_x \approx 10.4\ \text{m/s} \), and the vertical component is \( v_y \approx 6\ \text{m/s} \).

26. A golf ball is hit with an initial speed of 50 m/s at an angle of 15°. What is the time of flight?

Solution: Using the time of flight equation, the time is \( t \approx 2.6\ \text{s} \).

27. A projectile is fired from the ground with a speed of 40 m/s at 60° above the horizontal. What is its initial horizontal velocity component?

Solution: The horizontal component is \( v_x = v \cos \theta \approx 20\ \text{m/s} \).