# Down motion in free fall – problems and solutions

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

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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

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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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