A ball is dropped from a height of 20 m and rebounds with a velocity which is three fourth of the velocity with which it hits the ground
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The velocity of the ball just when it touches the ground can be calculated using the formula v² = u² + 2as, where v, u, a, s are the final velocity, initial velocity (=0, as the ball is at rest before it is droped), acceleration (= 10 m/s²) and displacement (=20 m) of the object respectively. Substituting the knows values, we get
v² = 0² + 2x10x20
=> v = 20 m/s
Now as the ball rebounds after hitting the ground, its initial velocity (u) becomes 3/4 th, that is, 15 m/s. When the object reaches the highest point after rebound, its final velocity (v) becomes zero. So using the equation, v = u + at, where v (=0), u (=15 m/s), a (= -10 m/s²) and t is the time taken by the ball to reach the highest point, we get
0 = 15 + (-10)x t
=> t = 1.5 seconds.
Now, assuming that there is no loss in energy, the ball will take the exact same time (= 1.5 s) to hit the ground after reaching the highest point as it took to reach the highest point.
So, according to your question, the time interval
v² = 0² + 2x10x20
=> v = 20 m/s
Now as the ball rebounds after hitting the ground, its initial velocity (u) becomes 3/4 th, that is, 15 m/s. When the object reaches the highest point after rebound, its final velocity (v) becomes zero. So using the equation, v = u + at, where v (=0), u (=15 m/s), a (= -10 m/s²) and t is the time taken by the ball to reach the highest point, we get
0 = 15 + (-10)x t
=> t = 1.5 seconds.
Now, assuming that there is no loss in energy, the ball will take the exact same time (= 1.5 s) to hit the ground after reaching the highest point as it took to reach the highest point.
So, according to your question, the time interval
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