A ball starts from rest on a horizontal floor and is given an acceleration of 2 m/s2. It hits the wall at a distance d from it. In the last second before it hits the wall, it covers a distance of 15 m. Find the distance d.
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A ball starts from rest on a horizontal floor and is given an acceleration of 2 m/s2. It hits the wall at a distance d from it. In the last second before it hits the wall, it covers a distance of 15 m. Find the distance d.
Acceleration of the ball = 2 m / Sec²
Distance = d meter
Distance covered at the last second before it hits the wall = 15 m
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So, 15 = 2/2 ( 2×n-¹ )
=> 15 = 2n-1
=> 2n = 16
=> n = 8 sec
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Formula used :-
u= u+at
v²=2as
s= v²/2a
= 0+2×8
= 16 m / sec
Putting in third formula :
So , the distance covered by the ball is 64 meter .
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