A ball is dropped from the top of a 95 m building. With what speed does the ball hit the ground?
The kinetic energy when it hits the ground equals the initial potential energy.
M*g*H = (1/2)MV^2
V = sqrt(2 g H)
H is the initial height and g is the acceleration of gravity.
Aerodynamic friction has been neglected. The true velocity will be slightly less
To calculate the speed at which the ball hits the ground, we can use the equation for the final velocity of an object in free fall:
vf^2 = vi^2 + 2aΔy
Where:
- vf is the final velocity of the ball.
- vi is the initial velocity of the ball (which is 0 in this case, since it is dropped).
- a is the acceleration due to gravity (which is approximately 9.8 m/s^2 near the surface of the Earth).
- Δy is the change in height of the ball (which is the height of the building, 95 m, in this case).
Plugging in the values, we get:
vf^2 = 0^2 + 2 * 9.8 * 95
Simplifying:
vf^2 = 2 * 9.8 * 95
vf^2 = 1862
Taking the square root of both sides to find vf:
vf = √1862
vf ≈ 43.14 m/s (rounded to two decimal places)
So, the ball hits the ground with a speed of approximately 43.14 m/s.