Solved Examples on Projectile Motion
Example 1: Determine the vertical distance of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 40 m/s at an angle of 45°, and it hit the ground after 4 seconds. (g = 10 m/s2)
Solution:
Given data:
Angle (θ) = 45°
Time = 4 sec
Initial Velocity (u) = 40 m/s
Horizontal distance (x) = uxt
x = ucosθ t
x = 40 × cos 45° × 4 = 113.137 m
From the trajectory formula, we have,
y = x tanθ − [gx2/2u2cos2θ]
y = 113.137 × tan 45° − [10 × (113.137)2/2 × (40)2 × (cos 45°)2]
y = 113.137 × 1 − [10 × 12,799.981/2 × 1600 × (1/√2)2]
y = 113.137 − [1,27,999.81/1600]
y = 113.137 − 79.99
= 33.147 m
Hence, the vertical distance covered by the ball is 33.147 m.
Example 2: A bullet is fired from a gun with a velocity of 15 m/s at an angle of 60°. Find the equation for the path of a projectile using the trajectory formula. (g = 9.8 m/s2)
Solution:
Given data:
Initial Velocity (u) = 15 m/s
Angle (θ) = 60°
From the trajectory formula, we have,
y = x tanθ − [gx2/2(ucosθ)2]
y = x tan 60° − [9.8 × x2/2(15 × cos 60°)2]
y = x × (√3) − [9.8x2/2 × (225/4)]
y = √3x – 0.087x2
Thus, the equation of the trajectory of the projectile is y =√3x – 0.087x2.
Example 3: Determine the vertical distance of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 13 m/s at an angle of 30°, and it hit the ground after 5 seconds. (g = 9.8 m/s2)
Solution:
Given data:
Angle (θ) = 30°
Time = 5 sec
Initial Velocity (u) = 13 m/s
Horizontal distance (x) = uxt
x = ucosθ t
x = 13 × cos 30° × 5 = 56.29 m
From the trajectory formula, we have,
y = x tanθ − [gx2/2u2cos2θ]
y = 56.29 × tan 30° − [9.8 × (56.29)2/2 × (13)2 × (cos 30°)2]
y = 56.29 × 1/√3 − [9.8 × 3,168.5641/2 × 169 × (√3/2)2]
y = 32.49 − [31,051.928/253.5]
y = 32.49 − 122.49
= −89.99 m
Example 4: A boy threw a stone with a velocity of 9 m/s at an angle of 35°. Find the equation for the path of a projectile using the trajectory formula. (g = 10 m/s2)
Solution:
Given data:
Initial Velocity (u) = 9 m/s
Angle (θ) = 35°
From the trajectory formula, we have,
y = x tanθ − [gx2/2(ucosθ)2]
y = x tan 35° − [10 × x2/2(9 × cos 35°)2]
y = x × (0.7002) − [10x2/2 × (9 × 0.819)2]
y = (0.7002) x – (0.0920) x2
Thus, the equation of the trajectory of the projectile is y = (0.7002) x – (0.0920) x2.
Derivation of Equation of Trajectory of a Projectile Motion
An object that is launched into space with only gravity acting on it is known as a projectile. The main force that acts on a projectile is gravity. Though other forces like air resistance also act on a projectile, their impact on the projectile is minimal when compared to the gravitational force. An arrow released from a bow, the launching of missiles, a bullet fired from a gun, a javelin thrown by an athlete, a ball thrown into the air, etc. are some real-life examples of projectiles.