Question

A ball thrown upwards, returns to the thrower after 6 seconds, with what speed does it return to the thrower? [Take g = 10 m/s2]

Answer 1

Answer:

30m/sec

Explanation:

The ball will return to the thrower at the same speed with which it was thrown.

Let it was thrown with speed u

Given that after 6 sec it returns to thrower so it takes 3 sec to reach the heighest point and another 3 sec to reach to the thrower

At maximum height it's velocity (v) will be zero.

Applying First Law of motion:-

v=u+(-g)t

0=u-10×3

u=30m/sec

So ball returns to the thrower with a speed of 30m/sec.

Hope it helps you!!!!

Answer 2

Answer:

The speed with which the ball returns to the thrower is 30 m/s.

Explanation:

• The initial velocity(u) of the ball when it starts its fall back to the owner, from the top= o m/s

• The final velocity of the ball(v) when it reaches the owner= ? m/s

• Time taken= [1/2 of 6 sec = 3 sec] [because equal time is taken to reach the top most point(3 sec) and return back to the thrower (3 sec)]
• Acceleration due to gravity= 10 m/s^2 (given)

[v=u+at]-------The first Equation of Motion

= [v=u+gt]

Thus:-

--- v=0+10(3)

--- v=30 m/s

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Have a nice time.

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