Gravitation — Question 18

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.