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# A ball is dropped from a height of 2.2 m and rebounds to a height of 1.9 m above the floor. Assume the ball was in con- tact with the floor for 96 ms and determine the average acceleration (magnitude and direction) of the ball during contact with the floor.

Navjyot Kalra
6 years ago
Given:
Height from which the ball was dropped, h1 = 2.2 m.
Height to which the ball rises,h2 = 1.9m .
Acceleration due to gravity,g = 9.8 m/s2 .
Initial speed of ball during its downward journey,v0 = 0 m /s .
Time for which the ball was in contact, t = 96 ms.
First we need to calculate the speed of the ball just before and after it hits the ground.
Let us assume that final speed of the ball during its downward journey is v1, and under the action of gravity the ball accelerates with g (the initial speed of the ball being zero).
From the equation of kinematics, we have

Where v represents the final speed of the object under consideration, a is its acceleration, x the distance travelled by the object,v0 the initial speed and t the time for which the object is studied.
From the second equation, the time t is given as:

Therefore the final speed of the ball before it hits the ground is 6.5 m/s and the velocity vector of the ball points downwards.
When the ball moves upwards, at the highest point its final speed (say v2) is zero.
The initial speed of the ball (say v01), that is the speed just after the contact with the ground can be calculated using equation (1) as:

The height to which the ball rises is given by x and is equal to h2. However, when the ball rise upwards, the acceleration due to gravity decelerates its motion, therefore the acceleration a in above equation is equated to -g.
Substitute the values in the equation above to have

Therefore the speed of the ball after the contact with the ground is 6.1 m/s. The velocity vector of the ball after the contact points in the upward direction.
Assume that the unit vector points along positive y axis such that the motion of the ball is restricted to y axis only, then the velocity vector of the ball before it hits the ground can be given as:

The negative sign highlights the fact that the velocity vector points downwards, opposite to the direction of unit vector .
The velocity vector after the ball has hit the ground and the contact is released is given as:

The positive sign highlights the fact that the velocity vector points in the direction of unit vector .

The magnitude of the acceleration vector in two significant figures is 130 m /s2.
Therefore the acceleration vector points in the direction of unit vector that is upward, and the magnitude of the acceleration is 130 m/s2.