A basketball player is running at 5.10 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 0.800 meters above the floor
v(vert) = sqrt (2gh) = =sqrt(2•9.8•0.8)=3.96 m/s
Use kinematic equation
v^2-u^2=2as
v=0 (at 0.8m above floor)
u=unknown
a=-9.81
s=0.8
u^2=√(2*9.81*0.8)=15.696
u=4 m/s approx.
To find the vertical velocity needed for the basketball player to rise 0.800 meters above the floor, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (which we are trying to find)
- vi is the initial velocity
- a is the acceleration (which is equal to the acceleration due to gravity, approximately 9.8 m/s^2)
- d is the displacement
In this case, the initial velocity in the vertical direction (vi) is 0 m/s since the player starts from rest in the vertical direction. The displacement (d) is 0.800 meters, and the acceleration (a) is 9.8 m/s^2.
Rearranging the equation, we get:
vf^2 = 2ad
Now, we can plug in the values:
vf^2 = 2 * 9.8 m/s^2 * 0.800 m
vf^2 = 15.68 m^2/s^2
Taking the square root of both sides, we find:
vf ≈ 3.96 m/s
Therefore, the basketball player needs a vertical velocity of approximately 3.96 m/s to rise 0.800 meters above the floor.