Question

A basketball player standing under the hoop shoots the ball straight up with an initial velocity of v0Part (a) What is the maximum height, h (in meters), above the launch point the basketball will achieve?

Answer 1

Answer:

Maximum height will be $h=\sqrt{\frac{v_0^2}{2g}}$

Explanation:

We have given initial velocity through which basketball is thrown $=v_0$

Acceleration due to gravity $=g\ m/sec^2$

At maximum height velocity will be zero

So final velocity v = 0 m /sec

According to third equation of motion $v^2=u^2+2gh$

$0^2=v_0^2-2gh$ ( Negative sign is due to upward acceleration )

$h=\sqrt{\frac{v_0^2}{2g}}$