Question

A basketball player jumps straight upward. After 0,625 s, she is 0.441 m above the ground. What is her initial velocity?????????

Answer 1

v₀ = 3.77 [m/s]

Step-by-step explanation:

This problem can be solved in a simple way by means of the following equation of kinematics.

`y=y_(o)+v_(o)*t+(1)/(2)*g*t^(2)`

where:

y - yo = 0.441 [m]

Vo = initial velocity [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time = 0.625 [s]

`0.441 = v_(o)*(0.625)-(1)/(2) *9.81*(0.625)^(2) \\2.357 = v_(o)*0.625\\v_(o)=3.77[m/s]`

Note: The sign of the acceleration is negative since the movement of the basketball player is against of the gravity acceleration.