Gravitation — Question 13

Given,

Initial velocity (u) = 49 ms^-1

Final velocity v at maximum height = 0

Acceleration due to earth gravity g = -9.8 ms^-2 (negative as ball is thrown up).

Let H be the maximum height to which the ball rises.

By third equation of motion,

2gH = v² - u²

Substituting we get,

2 × (-9.8) × H = 0 - (49)²

-19.6 H = -2401

H = $\dfrac{-2401}{-19.6}$ = 122.5 m

Hence, the maximum height to which it rises = 122.5 m

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

According to the first equation of motion,

v = u + gt

Substituting we get,

0 = 49 + (-9.8) x T_a

T_a = $\dfrac{49}{9.8}$ = 5 s

Also, T_d = 5 s

∴ T = T_a + T_d = 5 + 5 = 10 s

Hence, total time it takes to return to the surface of the earth = 10 s