A tennis ball is dropped from 1.93 m above the ground. It rebounds to a height of 0.921 m.With what velocity does it hit the ground?The acceleration of gravity is 9.8 m/s2.(Letdown be negative.)Answer in units of m/s.
a = -9.8
v = Vinitial - 9.8 t = -9.8 t
z = Zinitial + Vinitial t - 4.9 t^2 = 1.93 - 4.9 t^2
at ground z = 0
so
4.9 t^2 = 1.93
t = sqrt (1.93/4.9)
so
v at ground = -9.8 sqrt (1.93/4.9)
alternatively
kinetic energy at ground = loss in potential energy
(1/2) m |v|^2 = m g h
|v| = sqrt (2 g h)
|v| = sqrt (2 * 9.8 *1.93)
= 6.15 down (negative)
same answer
V^2 = Vo^2 + 2g*h = 0 + 19.6*1.93 = 37.8,
V = 6.15 m/s.
To find the velocity at which the tennis ball hits the ground, we can use the equation for vertical motion under constant acceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (which we want to find)
u = initial velocity (when the ball is dropped, u = 0)
a = acceleration due to gravity (a = -9.8 m/s^2, since we take downward direction as negative)
s = displacement (the distance the ball travels from the drop point to the point where it hits the ground, s = 1.93 m)
In this case, we want to find the final velocity (v) when the ball hits the ground. So, we rearrange the equation as follows:
v^2 = 0^2 + 2(-9.8)(1.93)
Simplifying this equation gives:
v^2 = -37.912
To solve for v, we take the square root of both sides:
v = √(-37.912)
Since the velocity should be a positive value (since we consider downward motion as negative), we can conclude that the ball hits the ground with a velocity of approximately 6.15 m/s (rounded to two decimal places).