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A ball is dropped vertically from a height $d$ above the ground. It hits the ground and bounces up vertically to a height $d/2$. Neglecting subsequent motion and air resistance, its velocity $v$ varies with the height $h$ above the ground is




Solution
(a)For the given condition initial height $h = d$ and velocity of the ball is zero.
When the ball moves downward its velocity increases and it will be maximum when the ball hits the ground & just after the collision it becomes half and in opposite direction.
As the ball moves upward its velocity again decreases and becomes zero at height $d/2$.
This explanation match with graph (A).
Similar Questions
A balloon rises, up with constant net acceleration of $10\,m / s ^2$. After $2\,s$ a particle drops from the balloon. After further $2\,s$ match the following columns. (Take $\left.g=10\,m / s ^2\right)$
colum $I$ | colum $II$ |
$(A)$ Height of particle from ground | $(p)$ $0$ |
$(B)$ Speed of particle | $(q)$ $10$ SI unit |
$(C)$ Displacement of particle | $(r)$ $40$ SI unit |
$(D)$ Acceleration of particle | $(s)$ $20$ SI unit |