Answer :
Answer:
The maximum height is [tex]h_{max} = \frac{v^2}{2g}[/tex]
Explanation:
From the question we are given that
[tex]K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,[/tex]
and [tex]K=(1/2)mv^2[/tex]
while [tex]U=mgh[/tex]
Now at the minimum height the kinetic energy is maximum and the potential energy is 0
[tex]K_i \ is \ max[/tex]
[tex]U_i = 0[/tex]
At maximum height [tex]h_{max}[/tex]
The kinetic energy is 0 and kinetic energy is
Hence the above equation
[tex]K_i = U_f[/tex]
[tex]\frac{1}{2}mv^2 = mgh_{max}[/tex]
Making [tex]h_{max}[/tex] the subject of the formula we have
[tex]h_{max} = \frac{v^2}{2g}[/tex]