Question

A ball thrown up vertically returns to the thrower after 6 s. Find. the velocity with which it was thrown up.

Answer 1

Consider, ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪  upward gravity = -9.8 m/s2   ● Total time (to and fro)= 6 s,   ●so, upward journey = 6/2 = 3 s ● Initial velocity (u) = ?   ●Final velocity (v) = 0 m/s   ●From First equation of motion,   ●Final velocity = Initial velocity +gt   •or, 0 = u + (-9.8 X3)  or, u = 29.4 m/s ▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪▪ So, The velocity at which ball is thrown in the upward direction is 29.4 m/s.

Answer 2

Dear Student,

◆ Answer -

u = 29.4 m/s

● Explanation -

Time required for ball to reach maximum height and time to return to throwers hand is same.

Hence, ball must have taken 6/2 = 3 s to reach maximum height.

By Newton's 1st law of motion,

v = u + at

Here, v = 0 m/s, a = -g, t = 3 s,

0 = u + (-9.8)×3

u = 29.4 m/s

Therefore, ball was thrown up with velocity of 29.4 m/s.

Thanks dear...