A ball of mass 100g is dropped from a height of 1.25m above the ground surface.It rebounds to a height of 1.1m.Find the velocity of the ball just before impact
KE at ground= PE at top
1/2 m v^2=m*g*1.25
solve for velocity v. Note that the mass of the ball divides out.
V^2 = Vo^2+2g*h = 0+19.6*1.25 = 24.5
V = 4.95 m/s.
To find the velocity of the ball just before impact, we can use the principle of conservation of energy.
The potential energy of the ball at the initial height is given by:
PE_initial = mgh
where m is the mass of the ball (in kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height (1.25 m).
The potential energy of the ball at the rebound height is given by:
PE_rebound = mgh'
where h' is the rebound height (1.1 m).
According to the conservation of energy, the total mechanical energy (potential energy + kinetic energy) of the ball is conserved throughout the motion, assuming no energy losses due to air resistance or other factors.
Therefore, we can equate the initial and rebound potential energies:
PE_initial = PE_rebound
mgh = mgh'
Simplifying the equation and canceling out the mass:
gh = gh'
Now, we need to find the final velocity of the ball just before impact. The final potential energy of the ball is given by:
PE_final = mgh_final
Since the ball is at ground level just before impact, h_final = 0.
Using the equation above, we rearrange it to find the final velocity:
gh_final = PE_final
v_final = √2gh_final
Plugging in the values, we have:
v_final = √(2gh)
Where v_final is the velocity of the ball just before impact.
Now, let's calculate the velocity.
Given:
m = 100 g = 0.1 kg
h = 1.25 m
First, calculate the potential energy at the rebound height:
PE_rebound = mgh' = (0.1 kg)(9.8 m/s²)(1.1 m) = 0.1078 J
Next, find the final velocity:
v_final = √(2gh) = √(2 * 9.8 m/s² * 1.25 m) = √(24.5) ≈ 4.95 m/s
Therefore, the velocity of the ball just before impact is approximately 4.95 m/s.