A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) The maximum height to which it rises, (ii) The total time it takes to return to the surface of the earth.

Solution:

Given data:

Initial velocity u = 49 m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s² (thus negative as ball is thrown up).

By third equation of motion,

2gH = v² – u²

2 × (- 9.8) × H = 0 – (49)²

– 19.6 H = – 2401

H = 122.5 m

Total time T = Time to ascend (T_a) + Time to descend (T_d)

v = u + gt

0 = 49 + (-9.8) x T_a

Ta = (49/9.8) = 5 s

Also, T_d = 5 s

Therefore T = T_a + T_d

T = 5 + 5

T = 10 s