A 0.50 kg softball is thrown vertically with a velocity of 30 m/s. What is the maximum height that the ball will reach?46 m15 m7.5 m92 m

Respuesta :

The maximum height reached by the ball is 46 m.

Given data:

The mass of the ball is m=0.50 kg.

The velocity of the ball is v=30 m/s.

The gravitational acceleration acting on the ball is g=9.8 m/s².

The maximum height that the ball reach will be,

[tex]\begin{gathered} h_{max}=\frac{v^2}{2g} \\ h_{max}=\frac{(30)^2}{2(9.8)} \\ h_{max}=46\text{ m} \end{gathered}[/tex]

Thus, the maximum height reached by the ball is 46 m.