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Question
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
- the maximum height to which it rises.
- the total time it takes to return to the surface of the earth.
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Solution
i. Initial velocity = 49m/s
Final velocity at maximum height = 0
g = −9.8m/s2 since the ball is moving in the opposite direction
Let the maximum height = h
Third equation of motion
v2 = u2 + 2gh
02 = (49)2 + 2x (−9.8) × h
2401 = 2 × 9.8h
122.5m = h
ii. Let the time taken to reach the maximum height = t
The first equation of motion
v = u + gt
0 = 49 + (−9.8) × t
49 = 9.8 × t
t = `49/9.8`
t = 5s
Time taken to reach maximum height = 5s
Total time taken by the ball to reach back to the earth 5 + 5 = 10s
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