# Determine the position of an object in projectile motion

1. A body is projected upward at an angle of 60o to the horizontal with an initial speed of 12 m/s. Determine the position of the object after moving for 1 second! Acceleration of gravity is 10 m/s2.

Known :

Angle (θ) = 60o

Initial velocity (vo) = 12 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s2

Wanted : Object position after moving during 1 second

Solution :

Horizontal component of initial velocity :

vox = vo cos θ = (12 m/s)(cos 60o) = (12 m/s)(0.5) = 6 m/s

Vertical component of initial velocity :

voy = vo sin θ = (12 m/s)(sin 60o) = (12 m/s)(0.53) = 63 m/s

Object position at horizontal direction:

Known :

Horizontal component of velocity (vx) = 6 m/s

Time interval (t) = 1 second

Wanted : horizontal range (x)

Solution :

6 meters / second means the ball moves as far as 6 meters every 1 second. The distance of the ball after moving for 1 second is 6 meters. So the position of the ball in the horizontal direction is 6 meters.

Object position at vertical direction :

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (vo) = 63 m/s (positive upward)

Time interval (t) = 1 second

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

Wanted : height after moving during 1 second

Solution :

h = vo t + 1/2 g t2 = (63)(1) + 1/2 (-10)(12) = 63 + (-5)(1) = 63 – 5 = 6(1.7) – 5 = 10.2 – 5 = 5.2 meters.

Object position after moving for 1 second :

Horizontal displacement (x) = 6 meters

Vertical displacement (y) = 5.2 meters

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2. A body is projected upward at an angle of 30o to the horizontal from a building 20 meters high. Its initial speed is 50 m/s. Calculate the vertical displacement after the body moving for 1 second! Acceleration of gravity is 10 m/s2.

Known :

Angle (θ) = 30o

Initial height (ho) = 20 meters

Initial velocity (vo) = 50 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s2

Wanted : Height (h)

Solution :

Vertical component of initial velocity :

voy = vo sin θ = (50 m/s)(sin 30o) = (50 m/s)(0.5) = 25 m/s

Height :

Choose upward direction as positive and downward direction as negative.

Known :

Initial velocity (vo) = 25 m/s (positive upward)

Time interval (t) = 1 second

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

Wanted : Height (h)

Solution :

h = vo t + 1/2 g t2 = (25)(1) + 1/2 (-10)(12) = 25 + (-5)(1) = 25 – 5 = 20 meters.

The height of the body after moving for 1 second is 20 meters above where the body is projected or 40 meters above ground.

3. A small ball projected horizontally with initial velocity vo = 10 m/s from a building 10 meters high. Calculate the ball displacement after moving 1 second! Acceleration of gravity is 10 m/s2

Known :

Initial height (h) = 10 meters

Initial velocity (vo) = 10 m/s

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s2

Wanted: Position of the ball after moving 1 second!

Solution :

Horizontal displacement :

Known :

Horizontal component of velocity (vx) = 10 m/s

Time interval (t) = 1 second

Wanted: Position of the object

Solution :

10 meters/second means the object move as far as 10 meters every 1 second. Displacement after moving for 1 second is 10 meters. So horizontal displacement is 10 meters.

Vertical displacement :

Calculated as the free fall motion.

Known :

Time interval (t) = 1 second

Acceleration of gravity (g) = 10 m/s2

Wanted : Height after moving during 1 second (h)

Solution :

h = 1/2 g t2 = 1/2 (10)(12) = (5)(1) = 5 meters.

After 1 second, the object falls as far as 5 meters. Height above the ground level = 10 meters – 5 meters = 5 meters.

The position of the object after moving 1 second :

Position of the object at horizontal direction (x) = 10 meters

The position of the object at vertical direction (y) = 5 meters