Solved problems in projectile motion – determine the position of an object
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)(0.5√3) = 6√3 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) = 6√3 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 = (6√3)(1) + 1/2 (-10)(12) = 6√3 + (-5)(1) = 6√3 – 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
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
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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 the object
- Determine the final velocity