An apple falls from a tree and hits the ground
10.9 m below.
With what speed will it hit the ground?
The acceleration of gravity is 9.8 m/s
2
.
Answer in units of m/s.
1/2 m v^2 = m g h ... v = √(2 g h)
v = √(2 * 9.8 * 10.9)
14.61
Well, let's get down to the core of the problem, shall we? When an apple falls from a tree, it's sure to make an impact, both literally and figuratively! To calculate the speed with which it hits the ground, we can make use of the basic kinematic equation.
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (which we assume to be zero since the apple is just falling), a is the acceleration due to gravity (-9.8 m/s^2, as it acts downwards), and s is the displacement (10.9 m), we can solve for v.
Plugging in the values, we can calculate v:
v^2 = 0 + 2(-9.8 m/s^2)(10.9 m)
v^2 = -213.08 m^2/s^2
But hold your horses, we can't have a negative velocity here! So we'll take the square root of the positive value:
v ≈ √213.08 ≈ 14.60 m/s
So, the apple hits the ground with a speed of approximately 14.60 m/s.
To find the speed at which the apple hits the ground, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (speed at which it hits the ground)
u is the initial velocity (which is 0 as the apple starts from rest when it falls)
a is the acceleration due to gravity (-9.8 m/s^2 as it acts downwards)
s is the vertical distance traveled by the apple (10.9 m)
Plugging in these values, we get:
v^2 = 0^2 + 2 * (-9.8 m/s^2) * 10.9 m
Simplifying the equation:
v^2 = 2 * (-9.8 m/s^2) * 10.9 m
v^2 = -2 * (9.8 m/s^2) * 10.9 m
v^2 = -213.68 m^2/s^2
Taking the square root of both sides to find v:
v = √(-213.68 m^2/s^2)
Since the velocity cannot be negative in this context, the apple will hit the ground with a speed of 14.6 m/s.