Solved problems in projectile motion – determine the horizontal displacement
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)(0.5√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 (vo) = 8√3 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 = (8√3) t + 1/2 (-10) t2
0 = 8√3 t – 5 t2
8√3 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.
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)(0.5√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 (vo) = 15√3 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 = (15√3) t + 1/2 (-10) t2
-50 = 15√3 t – 5 t2
5 t2 – 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.
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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- 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
