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) = u_xt 

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) = u_xt

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.

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.