A tennis ball is dropped from 1.98 m above
the ground. It rebounds to a height of 1.05 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s2
. (Let
down be negative.)
Answer in units of m/s
Confused please help.
v = -sqrt(2*9.8*1.98)
To find the velocity at which the tennis ball hits the ground, you can use the law of conservation of energy. According to this principle, the total mechanical energy of an object remains constant if no external forces are acting on it.
The mechanical energy of an object can be divided into two types: kinetic energy (KE) and potential energy (PE). When the ball is at its highest point (1.98 m above the ground), it has only potential energy. When it hits the ground, it has only kinetic energy.
Since the ball goes from a height of 1.98 m to a height of 0 m, the change in potential energy is given by:
ΔPE = m * g * Δh
Where ΔPE is the change in potential energy, m is the mass of the ball, g is the acceleration due to gravity, and Δh is the change in height.
At its highest point, the potential energy is entirely converted to kinetic energy:
PE = KE
Therefore:
m * g * Δh = 1/2 * m * v^2
Where v is the velocity at which the ball hits the ground. We can solve for v:
2 * g * Δh = v^2
Now let's substitute the given values into the equation:
2 * 9.8 m/s^2 * (1.98 m - 1.05 m) = v^2
Simplifying the equation gives:
v^2 = 19.6 m^2/s^2 * 0.93 m
v^2 = 18.168 m^2/s^2
Finally, taking the square root of both sides of the equation gives us the velocity:
v ≈ 4.263 m/s
Therefore, the velocity at which the tennis ball hits the ground is approximately 4.263 m/s.