An apple falls from a tree and hits the ground 9.63m below. With what speed will it hit the ground? The acceleration of gravity is 9.8 meters per second squared.
v = sqrt(2 g h)
because
(1/2) m v^2 = m g h
gain in kinetic energy = loss in potential energy
To find the speed at which the apple hits the ground, we can use the equation for motion under constant acceleration.
The equation we can use is:
v^2 = u^2 + 2as
Where:
v = final velocity (speed at which the apple hits the ground)
u = initial velocity (which in this case is 0, since the apple starts from rest)
a = acceleration due to gravity (which is 9.8 m/s^2 in this case)
s = displacement (the distance the apple falls, which is 9.63 m)
Plugging in the values into the equation, we get:
v^2 = 0^2 + 2 * 9.8 * 9.63
Simplifying further, we have:
v^2 = 2 * 9.8 * 9.63
v^2 = 189.42
To find the speed(v), we need to take the square root of both sides:
v = √189.42
v ≈ 13.77 m/s
Therefore, the speed at which the apple hits the ground is approximately 13.77 m/s.