A basketball player is running at$\mathbf{5}\mathbf{.}\mathbf{00}$m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise$\mathbf{0}\mathbf{.}\mathbf{750}$m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

When gravity first exerts force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

The initial velocity in y direction we need to calculate.

The time is also not given.

The displacement in y direction is$0.75$meter.

The gravitational acceleration is -$9.8$m/s²

The final velocity is also known to be 0.

(a) The velocity in the x frame will be constant in all time. Considering the initial velocity

$\begin{array}{rcl}{V_{fy}}^2& =& {V_{iy}}^2+2aY\\ 0^2& =& {V_{iy}}^2+2\left(-9.8\right)\left(0.75\right)\\ {V_{iy}}^2& =& 14.7\\ V_{iy}& =& \pm 3.91 m/s\\ & & \end{array}$

The initial velocity will be positive, that is $3.83$m/s.

(b) The time can be calculated by following equation

$\begin{array}{rcl}{V}_f& =& V_i+at\\ 0& =& 3.87+\left(-9.8\right)t\\ t& =& 0.391 s\\ & & \end{array}$

Hence the time taken is$0.391$S.

He should begin his jump in below distance

$\begin{array}{rcl}\overline{V}& =& \frac{x}{t}\\ 5& =& \frac{x}{0.391}\\ x& =& 1.96 meter\\ & & \end{array}$

He should begin his jump in below distance of $1.96$meter.