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 (v_o) = 10 m/s

Time elapsed (t) = 2 seconds

Acceleration of gravity (g) = 10 m/s²

Wanted : Final velocity (v_t)

Solution :

Acceleration 10 m/s²means 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 :

v_t = v_o + a t ………. 1

h = v_o t + ½ a t²………. 2

v_t² = v_o² + 2 a h ………. 3

v_t = v_o + g t

v_t = 10 + (10)(2)

v_t = 10 + 20 = 30 m/s

Final velocity = v_t = 30 m/s

[irp]

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 (v_o) = 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 = v_o t + ½ g t²

h = (5)(2) + ½ (10)(2)²

h = 10 + (5)(4)

h = 10 + 20

h = 30 meters

[irp]

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 (v_o) = 10 m/s

Acceleration of gravity (g) = 10 m/s²

Wanted :

(a) Time interval (t)

(b) Final velocity (v_t)

Solution :

(a) Time interval (t)

Final velocity :

v_t² = v_o² + 2 g h

v_t² = (10)² + 2(10)(80) = 100 + 1600 = 1700

v_t= 41 m/s

Time interval (t) :

v_t = v_o + g t

41 = 10 + (10)(t)

41 – 10 = 10 t

31 = 10 t

t = 31 / 10 = 3,1 seconds

(b) Final velocity (v_t₎ ?

v_t = 41 m/s

[irp]

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