Question

A ball thrown up vertically returns to the thrower after 6 s. Find

1. the velocity with which it was thrown up,
2. the maximum height it reaches, and
3. its position after 4 s.

Answer 1

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.