Solved problems in projectile motion – determine the horizontal displacement

1. A kicked football leaves the ground at an angle θ = 60^{o }with the horizontal has an initial speed of 16 m/s. How long will it be before the ball hits the ground?

__Known :__

Angle (θ) = 60^{o}

Initial speed (v_{o}) = 16 m/s

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

__Wanted :__ Horizontal displacement (x)

__Solution :__

Horizontal component of initial velocity :

v_{ox }= v_{o} cos θ = (16 m/s)(cos 60^{o}) = (16 m/s)(0.5) = 8 m/s

Vertical component of initial velocity :

v_{oy }= v_{o} sin θ = (16 m/s)(sin 60^{o}) = (16 m/s)(0.5√3) = 8√3 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**

The time it stays in the air is determined by the y motion. We first find the time using the y motion and then use this time value in the x equations (constant velocity equation).

Choose upward direction as positive and downward direction as negative.

__Known :__

Initial velocity (v_{o}) = 8√3 m/s (v_{o} upward)

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

Height (h) = 0 (ball is back to the same position)

__Wanted :__ Time in air

__Solution :
__

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

0 = (8√3) t + 1/2 (-10) t^{2}

0 = 8√3 t – 5 t^{2}

8√3 t = 5 t^{2}

8 (1.7) = 5 t

14 = 5 t

t = 14 / 5 = 2.8 seconds

**Horizontal displacement**

__Known :__

Velocity (v) = 8 m/s

Time interval (t) = 2.8 seconds

__Wanted :__ Displacement

__Solution :
__

x = v t = (8 m/s)(2.8 s) = 22.4 meters

Horizontal displacement is 22.4 meters.

[irp]

2. A body is projected upward at angle of 60^{o }with the horizontal from a building 50 meter high. It’s initial speed is 30 m/s. Calculate the horizontal displacement! Acceleration of gravity is 10 m/s^{2}.

__Known :__

Angle (θ) = 60^{o}

High (h) = 15 m

Initial speed (v_{o}) = 30 m/s

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

__Wanted :__ x

__Solution :__

Horizontal component of initial velocity ::

v_{ox }= v_{o} cos θ = (30 m/s)(cos 60^{o}) = (30 m/s)(0.5) = 15 m/s

Vertical component of initial velocity :

v_{oy }= v_{o} sin θ = (30 m/s)(sin 60^{o}) = (30 m/s)(0.5√3) = 15√3 m/s

**Time in the air**

We first find the time using the y motion and then use this time value in the x equations (constant velocity equation). Choose upward as positive and downward as negative.

__Known :__

Initial velocity (v_{o}) = 15√3 m/s (positive upward)

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

High (h) = -50 (Ground 50 meter below the initial position)

__Wanted :__ t

__Solution :__

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

-50 = (15√3) t + 1/2 (-10) t^{2}

-50 = 15√3 t – 5 t^{2}

5 t^{2} – 15√3 t – 50 = 0

Calculate time using this formula :

a = 5, b = –15√3, c = –50

Time in the air is 6.7 seconds.

**Horizontal displacement :**

__Known :__

Velocity (v) = 15 m/s

Time interval (t) = 6.7 seconds

__Wanted :__ displacement

__Solution :__

s = v t = (15 m/s)(6.7 s) = 100.5 meters

Horizontal displacement is 100.5 meters.

[irp]

3. A small ball projected horizontally with initial velocity v_{o} = 10 m/s from a building 10 meter high. Calculate the horizontal displacement! Acceleration of gravity is 10 m/s^{2}

__Known :__

High (h) = 10 m

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

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

__Wanted :__ x

__Solution :__

Horizontal component of initial velocity = initial velocity = 10 m/s.

**Time in the air**

Time in air calculated using free fall motion equation.

__Known :__

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

High (h) = 10 meter

__Wanted :__ t

__Solution :__

h = 1/2 g t^{2}

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

10 = 5 t^{2}

t^{2 }= 10 / 5 = 2

t = √2 = 1.4 seconds

**Horizontal displacement**

Horizontal displacement calculated using equation of motion at constant velocity.

__Known :__

Velocity (v) = 10 m/s

Time interval (t) = 1.4 seconds

__Wanted :__ x

__Solution :__

s = v t = (10 m/s)(1.4 s) = 14 meters

Horizontal displacement is 14 meters.

[irp]

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