From a height of 20m above a horizontal floor, a ball is thrown down with initial velocity of 20 m/s. After striking the floor, the ball bounces to the same height from which it was thrown. Find the coefficient of restitution for the collision between the ball and the floor. (g = 10 m/s²)
Answers
ball is thrown from a height , h = 20m
velocity of ball to reach the floor , v² = u² + 2as
v² = 20² + 2 × 10 × 20
v = 20√2 m/s
so, velocity of ball before collision = 20√2 m/s
a/c to question,
the ball bounces to the same height from which it was thrown.
so, at a 20m height , velocity of ball , v= 0
we have to find velocity of ball after collision
so, use formula, v² = u² + 2as
0 = u² + 2(-10) × 20
v = 20m/s
In the case of a ball bouncing off a flat, stationary surface, the coefficient of restitution turns out to be:
coefficient of restitution = V/U
where V is velocity of object after collision.
U is velocity of object before collision.
hence, coefficient of restitution = 20/20√2 = 1/√2
Answer:1/√2
Explanation:initial velocity of ball, u = 20m/s [ downward]
ball is thrown from a height , h = 20m
velocity of ball to reach the floor , v² = u² + 2as
v² = 20² + 2 × 10 × 20
v = 20√2 m/s
so, velocity of ball before collision = 20√2 m/s
a/c to question,
the ball bounces to the same height from which it was thrown.
so, at a 20m height , velocity of ball , v= 0
we have to find velocity of ball after collision
so, use formula, v² = u² + 2as
0 = u² + 2(-10) × 20
v = 20m/s
In the case of a ball bouncing off a flat, stationary surface, the coefficient of restitution turns out to be:
coefficient of restitution = V/U
where V is velocity of object after collision.
U is velocity of object before collision, coefficient of restitution = 20/20√2 = 1/√2