Solved Problems in Linear Motion – Down motion in free fall
1. A ball is thrown vertically downward with initial speed 10 m/s and reach the ground in 2 seconds. Find final speed just before the ball hits the ground. Acceleration of gravity (g) = 10 m/s2. Ignore air resistance.
Known :
Initial velocity (vo) = 10 m/s
Time elapsed (t) = 2 seconds
Acceleration of gravity (g) = 10 m/s2
Wanted : Final velocity (vt)
Solution :
Acceleration 10 m/s2 means speed increase by 10 m/s each second. After 3 second, speed = 30 m/s.
Final velocity = 10 m/s + 20 m/s = 30 m/s.
Kinematic equations for motion at constant acceleration, as shown below :
vt = vo + a t ………. 1
h = vo t + ½ a t2 ………. 2
vt2 = vo2 + 2 a h ………. 3
vt = vo + g t
vt = 10 + (10)(2)
vt = 10 + 20 = 30 m/s
Final velocity = vt = 30 m/s
2. A stone is thrown vertically downward from a bridge with initial speed 5 m/s and reach the water in 2 seconds. Calculate the height of the bridge.
Known :
Initial velocity (vo) = 5 m/s
Time elapsed (t) = 2 seconds
Acceleration due to gravity (g) = 10 m/s2
Wanted : the height of the bridge (h)
Solution :
h = vo t + ½ g t2
h = (5)(2) + ½ (10)(2)2
h = 10 + (5)(4)
h = 10 + 20
h = 30 meters
3. A ball is thrown vertically downward with initial speed 10 m/s from a height of 80 meters. Find (a) Time in air (b) Final velocity just before ball strikes the ground.
Known :
height (h) = 80 meters
Initial velocity (vo) = 10 m/s
Acceleration of gravity (g) = 10 m/s2
Wanted :
(a) Time interval (t)
(b) Final velocity (vt)
Solution :
(a) Time interval (t)
Final velocity :
vt2 = vo2 + 2 g h
vt2 = (10)2 + 2(10)(80) = 100 + 1600 = 1700
vt = 41 m/s
Time interval (t) :
vt = vo + g t
41 = 10 + (10)(t)
41 – 10 = 10 t
31 = 10 t
t = 31 / 10 = 3,1 seconds
(b) Final velocity (vt) ?
vt = 41 m/s
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- Distance and displacement
- Average speed and average velocity
- Constant velocity
- Constant acceleration
- Free fall motion
- Down motion in free fall
- Up and down motion in free fall
