How fast must a fall be thrown upward to reach a height of 12 m?

the initial kinetic energy converts to potential energy at the top.

KE=PE
1/2 m v^2=mgh

solve for v. Notice the mass of the "fall" divides out.

To determine the speed at which an object must be thrown upwards to reach a specific height, we can use the principles of projectile motion and apply the laws of kinematics.

The most fundamental equation of motion that we will use is:

v^2 = u^2 + 2as

where:
- v is the final velocity (or in this case, velocity when the object reaches its highest point),
- u is the initial velocity (or the velocity at which the object is thrown),
- a is the acceleration (which is equal to the acceleration due to gravity, approximately 9.8 m/s^2),
- s is the displacement, which in this case is the height the object needs to reach.

Since the object is thrown upward, the acceleration due to gravity will be acting in the opposite direction to the motion. Therefore, we take the acceleration as negative (-9.8 m/s^2).

In this case, we need to find the initial velocity (u).

Given:
Height (s) = 12 m
Acceleration (a) = -9.8 m/s^2

Rearranging the equation, we get:

v^2 = u^2 + 2as
u^2 = v^2 - 2as
u = sqrt(v^2 - 2as)

Since the object reaches its highest point when its final velocity (v) is 0, we can substitute v = 0 into the equation:

u = sqrt(0^2 - 2 * (-9.8) * 12)

Simplifying further:

u = sqrt(0 - (-235.2))
u = sqrt(235.2)
u ≈ 15.34 m/s

Therefore, the object must be thrown upwards with an initial velocity of approximately 15.34 m/s in order to reach a height of 12 m.

How do you throw a fall?