# Determine final velocity of projectile motion

1. A kicked football leaves the ground at an angle θ = 30^{o} to the horizontal with an initial speed of 14 m/s. Calculate the final velocity before the ball hits the ground.

__Known :__

Angle (θ) = 3^{o}

Initial velocity (v_{o}) = 14 m/s

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

__Wanted :__ Final velocity before the ball hits the ground

__Solution :__

Horizontal component of initial velocity :

v_{ox }= v_{o} cos θ = (14 m/s)(cos 30^{o}) = (14 m/s)(0.5√3) = 7√3 m/s

Vertical component of initial velocity :

v_{oy }= v_{o} sin θ = (14 m/s)(sin 30^{o}) = (14 m/s)(0.5) = 7 m/s

**Final velocity at vertical direction**

Choose upward direction as positive and downward direction as negative.

__Known :__

Initial velocity (v_{o}) = 7 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s^{2 }(negative downward)

Height (h) = 0 (object back to initial position)

__Wanted :__ Final velocity (v_{t})

__Solution :__

v_{t}^{2} = v_{o}^{2 }+ 2 g h = 7^{2 }+ 2(-10)(0) = 49 – 0 = 49

v_{t }= √49 = 7 m/s

**Final velocity at horizontal direction**

Initial velocity at horizontal direction is 7√3 m/s. Velocity is constant so that final velocity is same as initial velocity.

**Final velocity before the object hits the ground**

2. A body is projected upward at an angle of 30^{o }with the horizontal from a building 5 meter high. Its initial speed is 10 m/s. Calculate final velocity before the object hits the ground! Acceleration of gravity is 10 m/s^{2}.

__Known :__

Angle (θ) = 30^{o}

Initial height (h_{o}) = 5 meters

Initial velocity (v_{o}) = 10 m/s

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

__Wanted :__ Final velocity

__Solution :__

Horizontal component of initial velocity :

v_{ox }= v_{o} cos θ = (10 m/s)(cos 30^{o}) = (10 m/s)(0.5√3) = 5√3 m/s

Vertical component of initial velocity :

v_{oy }= v_{o} sin θ = (10 m/s)(sin 30^{o}) = (10 m/s)(0.5) = 5 m/s

**Final velocity at vertical direction**

__Known :__

Initial velocity (v_{o}) = 5 m/s (positive upward)

Acceleration of gravity (g) = –10 m/s^{2 }(negative downward)

Height (h) = -5 m (negative because the ground is below the initial height)

__Wanted :__ Final velocity (v_{t})

__Solution :__

v_{t}^{2} = v_{o}^{2 }+ 2 g h = 5^{2 }+ 2(-10)(-5) = 25 + 100 = 125

v_{t} = √125 m/s

**Final velocity at horizontal direction**

Final velocity at horizontal direction is 5√3 m/s.

**Final velocity **

3. A small ball projected horizontally with initial velocity v_{o} = 8 m/s from a building 12 meters high. Calculate final velocity before ball hits ground! Acceleration of gravity is 10 m/s^{2}

__Known :__

Height (h) = 12 meters

Initial velocity (v_{o}) = 8 m/s

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

__Wanted :__ Final velocity (v_{t})

__Solution :__

Horizontal component of initial velocity :

v_{ox }= v_{o} = 8 m/s

Vertical component of initial velocity :

v_{oy }= 0 m/s

**Final velocity at vertical direction**

calculated using equation of free fall motion.

__Known :__

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

Height (h) = 12 m

__Wanted :__ Final velocity (v_{t})

__Solution :__

v_{t}^{2} = 2 g h = 2(10)(12) = 240

v_{t }= √240 m/s

**Final velocity at horizontal direction**

Initial velocity at the horizontal direction is 8 m/s. Velocity is constant so that the initial velocity equals the final velocity. So final velocity at horizontal direction is 8 m/s.

**Final velocity **

- Resolve initial velocity into horizontal and vertical components
- Determine the horizontal displacement
- Determine the maximum height
- Determine the time interval
- Determine the position of the object
- Determine the final velocity