NCERT Solutions Class 9 Science Solutions for Gravitation - Exercise 10.6 in Chapter 10 - Gravitation
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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.
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
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