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Question
A ball thrown up vertically returns to the thrower after 6 s. Find
- the velocity with which it was thrown up,
- the maximum height it reaches, and
- its position after 4 s.
Solution
a. Final velocity (v) = 0
Time (t) = 6s
g = 10m/s2
The first equation of motion
v = u + gt
0 = u + (−9.8) × 3 (negative sign due to opposite direction)
u = 30m/s
Hence, the ball is thrown upwards with a velocity of 30m/s.
b. Initial velocity (u) = 30m/s
Final velocity (v) = 0
Time (t) = 3s
the second equation of motion
h = ut + `1/2` gt2
h = `30 × 3 + 1/2 xx (−10) x (3)^2`
h = 90 − 45
h = 45m
Hence, the maximum height attained by the ball is 45m.
c. Position of the ball after 4 s Distance travelled downwards in 1 sec the second equation of motion
h = ut + `1/2` gt2
h = `0 xx 1 + 1/2 xx 10 xx (1)^2`
h = 5m
Distance covered in 1 second = 5m
Distance covered in 4 seconds = 45 − 5
= 40 m
Hence, after 4 seconds, the ball will be 40m above the ground.
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