Question

A baseball is thrown vertically upward with an initial velocity of 30.0 m/s. The maximum height gained by the ball is?

Answer 1

Answer:

45.9 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the mechanical energy (sum of gravitational potential energy and kinetic energy) of the ball just after it is thrown is equal to the mechanical energy of the ball when it is at the top of its trajectory:

$E_i = E_f\\U_i + K_i = U_f + K_f$

where

$U_i = 0$ is the initial gravitational potential energy of the ball, which is zero because the ball starts its motion from the ground (h=0)

$K_i = \frac{1}{2}mv^2$ is the initial kinetic energy of the ball, where m is the mass of the ball and v=30.0 m/s is its speed

$U_f = mgh$ is the gravitational potential energy of the ball at its maximum height h

$K_f = 0$ is the kinetic energy of the ball when it is at the maximum height (it is zero because the speed of the ball at its maximum height is zero)

Substituting into the first equation, we can find h:

$\frac{1}{2}mv^2 = mgh\\h=\frac{v^2}{2g}=\frac{(30.0 m/s)^2}{2(9.8 m/s^2)}=45.9 m$