Gravitation | Gravitation - Exercise 10.6

Question

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.

Accepted Answer

Given data:

Initial velocity u = 49m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/ $s^2$ (thus negative as ball is thrown up).

By the third equation of motion,

$v^{2}=u^{2}-2 g s$

Substitute all the values in the above equation

$\begin{array}{l} 0=(49)^{2}-2 \times 9.8 \times s \\ S=\frac{(49)^{2}}{2 \times 9.8} \\ s=122.5 \mathrm{m} \end{array}$

Total time T = Time to ascend (Ta) + Time to descend (Td)

V = u – gt

0 = 49 – 9.8 x Ta

Ta = (49/9.8) = 5s

Also, Td = 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

Question 7