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

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4s.

Given,

Ball returns to thrower in 6s. So, time_up + time_down = 6 s hence, time_up = 3 s.

g = 9.8 ms^-2

(a) Final velocity (v) = 0

From first equation of motion

v = u – gt_up

⇒ u = v + gt_up

⇒ u = 0 + (9.8 x 3)

⇒ u = 29.4 ms^-1

Hence, the velocity with which it was thrown up = 29.4 ms^-1.

(b) According to the second equation of motion,

S = ut - $\dfrac{1}{2}$gt²

Substituting we get,

S = (29.4 x 3) - ($\dfrac{1}{2}$ x 9.8 x 3 x 3)
= 88.2 - 44.1 = 44.1 m

Hence, maximum height stone reaches is 44.1 m

(c) In 3 sec, it reaches the maximum height.

Distance travelled in another 1 sec = s'

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$gt²

g is +ve as ball is going down

Substituting we get,

S = 0 + ($\dfrac{1}{2}$ x 9.8 x 1 x 1)
= 4.9 m

Hence, in 4 sec the ball will be 4.9 m from the top or 39.2 m (i.e., 44.1 - 4.9) from the bottom.