Question

Problem 3: A basketball player is running at 4.75 m/s toward the basket when he jumps into the air to dunk the ball. He maintains the same horizontal velocity while in the air. Part (a) What vertical velocity does he need to rise 0.750 m above the floor in m/s? Numeric : A numeric value is expected and not an expression Voy = Part (b) At what horizontal distance from the basket, in meters, must he start his jump to reach his maximum height at the same time as he reaches the basket in m? Numeric : A numeric value is expected and not an expression.

Answer 1

Answer

given,

speed of the basketball player, v_x = 4.75 m/s

height of the jump = ?

a) initial vertical velocity = 0 m/s

   final vertical velocity = ?

       height, h = 0.75 m

 using equation of motion

   v² = u² +  2 g h

   v² = 0² +  2 x 9.8 x  0.75

   v² = 14.7

   v = 3.83 m/s

b) let horizontal distance= x m

 maximum height at time , t s

 time taken to reach maximum height

    $t = \dfrac{3.83}{9.8}$

         t = 0.39 s

    horizontal distance

             = v_x  × t

             = 4.75 × 0.39

             = 1.85 m

horizontal distance is equal to 1.85 m