If the tennis ball were in contact with the ground for 0.0108 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.
Answers
1.43 = 1/2gt^2
1.43/4.9 = t^2 = 0.29, sq-rt = t = 0.54 secs
v = u + gt ( since the ball was dropped we omit (u)
v = gt = 9.8 x 0.54 = - 5.29 m/s
b) With what velocity does it leave the ground? Answer in units of m/s.
2gh = v^2
2 x 9.8 x 0.801 = 15.6996, sq-rt = v = 3.9623 m/s
c) If the tennis ball were in contact with the ground for 0.00541 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.
v = at
3.9623 m/s = at
3.9623/t = 3.9623/0.00541 = 732.4 m/s^2
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Concept:
- Free fall
- One-dimensional motion
- Applying kinematics equations
Given:
- Height from which ball is released = 1.43 m
- Time of contact with ground = 0.0108 s
- Acceleration due to gravity g = 9.8 m/s^2
- Rebound height = 0.801 m
Find:
- The acceleration given to the tennis ball by the ground.
Solution:
We know the following kinematics equation,
v^2 = u^2 +2as
where v is the final velocity, when the ball hits the ground, u is the initial velocity which is 0, a =g, acceleration due to gravity and s is the initial height of the ball from where it has been released
v^2 = 0 +2(9.8)(1.43) = 28.028
v = 5.29 m/s
The velocity with which the ball leaves the ground is given by the conservation of energy
1/2mv^2 = mgh
v^2 = 2gh
v^2 = 2*9.8*0.801 = 15.6996
v = 3.96m/s
The change in velocity is 3.96 + 5.29 = 9.25 m/s
The acceleration is given by the change in velocity per unit time
a = 9.25/0.0108 = 856 m/s^2
The acceleration is 856m/s^2.
A tennis ball was dropped from a height of 1.43m. If the tennis ball were in contact with the ground for 0.0108 s and rebounds to a height 0.801 , find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.
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