A ball is thrown vertically upwards with a velocity of 49m/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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Solution
Given Initial velocity of ball, u=49m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Assumption: g=9.8m/s2 holds true (maximum height reached is small compared to the radius of the earth)
Velocity of the ball at maximum height is zero, v=0
Using the third equation of motion we get,
v2−u2=2aH
0−(49)2=2×(−9.8)×H
⟹H=122.5m
Using the first equation of motion we get,
v=u+at
0=49−9.8t
⟹t=5 s
∴ Total time taken by ball to return to the surface, T=2t=10s.