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]
Answers
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:
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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