Question

A ball is thrown up vertically returns to the thrower after 8s. Its heights above the ground after 6s will be ( take g= 10m/s^2)

Answer 1

Time of ascent is equal to the time of descent. The ball takes a total of 8 s for its upward and downward journey.
Hence, it has taken 4 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g=−10ms−2
From the first equation of motion,
v = u + gt
0=u+(−10×4)
u=10×4=40ms−1
Hence, the ball was thrown upwards with a velocity of 40 ms−1

Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u=40 ms−1
Final velocity, v = 0
Acceleration due to gravity, g=−10ms−2
From the second equation of motion,
s=ut+12gt2
h=40×4+12×−10×(4)2




=160-80=80m
Ball attains the maximum height after 4 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Distance travelled by it during its downward journey in 2 s is given by
s=0×t+12×10×(2)²=20 m
Total height = 80 m
This means that the ball is (80 m - 20 m)=60 m above the ground after 6 seconds.