Answer:
Maximum height will be [tex]h=\sqrt{\frac{v_0^2}{2g}}[/tex]
Explanation:
We have given initial velocity through which basketball is thrown [tex]=v_0[/tex]
Acceleration due to gravity [tex]=g\ m/sec^2[/tex]
At maximum height velocity will be zero
So final velocity v = 0 m /sec
According to third equation of motion [tex]v^2=u^2+2gh[/tex]
[tex]0^2=v_0^2-2gh[/tex] ( Negative sign is due to upward acceleration )
[tex]h=\sqrt{\frac{v_0^2}{2g}}[/tex]
