Manas Shukla
Last Activity: 8 Years ago
Taking gravity = 10m/s^2
Lets first find out the change in velocity of the ball while its in contact with the ground .
So , Let Velocity of the ball just before contact with the ground = v1
applying formula v^2=u^2 +2 a S
v1^2 = 0 + 2 x 10 x 10
v1 = root of (200) = 14.14
Let velocity of ball just after leaving the ground = v2
applying formula again v^2=u^2 +2 a S
0 = v2^2 – 2 x 10 x 2.5
v2 = root (50) = 7.07
Notice that v2 and v1 are in different directions
Now applying formula v = u + at
v2 = -v1 + a x .01
14.14 = -7.07 + .01 x a
1414 = -707 + a
acceleration = 1414 + 707 = 2121 m/s^2
