Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

1. the maximum height to which it rises.
2. the total time it takes to return to the surface of the earth.

Answer 1

i. Initial velocity = 49m/s

Final velocity at maximum height = 0

g = −9.8m/s² since the ball is moving in the opposite direction

Let the maximum height = h

Third equation of motion

v² = u² + 2gh

0² = (49)² + 2x (−9.8) × h 

2401 = 2 × 9.8h

122.5m = h

ii.  Let the time taken to reach the maximum height = t

The first equation of motion

v = u + gt 

0 = 49 + (−9.8) × t

49 = 9.8 × t

t = `49/9.8`

t = 5s

Time taken to reach maximum height = 5s

Total time taken by the ball to reach back to the earth 5 + 5 = 10s