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.

Asked by Shivani Kumari | 1 year ago |  237

1 Answer

Solution :-

Given data:

Initial velocity u = 49m/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,

v2 = u2 - 2gs

Substitute all the values in the above equation

\( 0=(49)^2-2\times 9.8 \times s\)

 \(S = \frac{(49)^2}{2 \times 9.8}\)

s = 122.5m

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

V = u - gt

 \(0 = 49 - 9.8 \times Ta\)

Ta = (49/9.8) = 5s

Also, Td 5s

Therefore T = Ta + Td

T = 5 + 5

T = 10s

 

Answered by Vishal kumar | 1 year ago

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