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07/01/2021 4:03 pm
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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.
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07/01/2021 4:24 pm
Initial velocity, u = 49 m/s
Acceleration, a = g = -9.8 m/s2
Velocity at the highest point, v = 0 m/s
(i) If h is the maximum height reached by the ball, then
2gs = \(v^2 - u^2\)
or s = \(\frac{v^2 - u^2}{2g}\) = \(\frac{(0 m/s)^2 - (49 m/s)^2}{2\times (-9.8 m/s^2)}\)
s = \(\frac{- (49 \times 49)m^2/s^2}{-(2 \times 9.8)m/s^2}\) = 122.5
Thus, the ball will reach a height of 122.5 m.
(ii) We have, v = u + gt
Here, 0 = 49 - 9.8 t
t = 49/9.8 = 5 s
Total time = 2t = 2 x 5s = 10 s