A basketball player can jump 2.16 m off the hardwood floor. With what upward velocity did he leave the floor? (use 9.8 m/s/s)
Try using the velocity formula:(V=velocity, d=distance, and t=time)
V=d/t
9.8(2.16/s)=
Ariel's answers won't be much help.
Use the fact that initial kientic energy in converted to potential energy (raising of the center of mass, by H = 2.16 m).
M g H = (1/2) M V^2
The M's cancel.
V = sqrt(2 g H) = 6.5 m/s
To find the upward velocity with which the basketball player left the floor, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at the highest point of the jump, as the ball momentarily stops)
u = initial velocity (what we want to find)
a = acceleration (-9.8 m/s^2, since gravity is acting in the opposite direction of the player's jump)
s = displacement (2.16 m, the height reached by the player)
Rearranging the equation, we get:
u^2 = v^2 - 2as
Since the player momentarily stops at the highest point, v = 0 m/s, so:
u^2 = 0^2 - 2(-9.8 m/s^2)(2.16 m)
u^2 = 0 - (-42.336 m^2/s^2)
u^2 = 42.336 m^2/s^2
Taking the square root of both sides, we find:
u ≈ √42.336 m/s
u ≈ 6.505 m/s
Therefore, the upward velocity with which the basketball player left the floor is approximately 6.505 m/s.