Basic Physics

# Determine horizontal displacement of projectile motion

1. A kicked football leaves the ground at an angle θ = 60o 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) = 60o

Initial speed (vo) = 16 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : Horizontal displacement (x) Solution :

Horizontal component of initial velocity :

vox = vo cos θ = (16 m/s)(cos 60o) = (16 m/s)(0.5) = 8 m/s

Vertical component of initial velocity :

voy = vo sin θ = (16 m/s)(sin 60o) = (16 m/s)(0.53) = 83 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 (vo) = 83 m/s (vo upward)

Acceleration of gravity (g) = -10 m/s2 (g downward)

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

Wanted : Time in air

Solution :

h = vo t + 1/2 g t2

0 = (83) t + 1/2 (-10) t2

0 = 83 t – 5 t2

83 t = 5 t2

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.

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2. A body is projected upward at angle of 60o 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/s2.

Known :

Angle (θ) = 60o

High (h) = 15 m

Initial speed (vo) = 30 m/s

Acceleration of gravity (g) = 10 m/s2

Wanted : x

Solution : Horizontal component of initial velocity ::

vox = vo cos θ = (30 m/s)(cos 60o) = (30 m/s)(0.5) = 15 m/s

Vertical component of initial velocity :

voy = vo sin θ = (30 m/s)(sin 60o) = (30 m/s)(0.53) = 153 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 (vo) = 153 m/s (positive upward)

Acceleration of gravity (g) = -10 m/s2 (negative downward)

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

Wanted : t

Solution :

h = vo t + 1/2 g t2

-50 = (153) t + 1/2 (-10) t2

-50 = 153 t – 5 t2

5 t2153 t – 50 = 0

Calculate time using this formula :

a = 5, b = –153, 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.

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3. A small ball projected horizontally with initial velocity vo = 10 m/s from a building 10 meter high. Calculate the horizontal displacement! Acceleration of gravity is 10 m/s2

Known :

High (h) = 10 m

Initial velocity (vo) = 10 m/s

Acceleration of gravity (g) = 10 m/s2

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/s2

High (h) = 10 meter

Wanted : t

Solution :

h = 1/2 g t2

10 = 1/2 (10) t2

10 = 5 t2

t2 = 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.

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