Question

a boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. the ball is 20 m above the ground as he releases it.

Answer 1

Answer: 25.1m

Explanation:

The expression of the initial potential energy when the ball is tossed upwards is given as followed

$PE_{i}$=mg$PE_{i}= mgh_{0}$

Here, m is the mass of the ball, g is the acceleration due to gravity and is the initial height.

The initial kinetic energy of the ball is:

$KE_{i} =\frac{1}{2}mv^2$

The final potential energy of the ball at the maximum height is given as:

$PE_{f}=mgh_{max}$

The final kinetic energy is at the maximum height is:

$KE_{f} =0$

Now you will use the conservation of energy princi$h_{0}$ple and substitute the equations into the expression below:

$KE_{i}+PE_i=KE_f+PE_f$

$\frac{1}{2}mv^2+mgh_0=0+mgh_{max}$

$h_{max}=h_0+\frac{v^2}{2g}$

Now substitute 9.81 m/$s^{2}$ for g, 10m/s for v, and 20.0 m for $h_{0}$ in the equation above.

h$_{max}}=(20.0m)+\frac{(10m/s)^2}{2(9.81m/s^2}$

which is equal to 25.1m