a basketball player can leap upward 0.5m. what is his initial velocity at the start of the leap?
Vf^2=Vi^2-2g*h
Vf=0, vi=?, h=.5m
solve for Vi
To determine the initial velocity of the basketball player at the start of the leap, we can use the kinematic equation:
vf^2 = vi^2 + 2as
Where:
vf = final velocity (0 m/s at the peak of the leap)
vi = initial velocity (unknown)
a = acceleration (the acceleration due to gravity, which is approximately -9.8 m/s^2)
s = displacement (0.5 m)
Since the final velocity at the peak of the leap is 0 m/s, we can rewrite the equation as:
0 = vi^2 + 2(-9.8)(0.5)
Simplifying:
0 = vi^2 - 9.8
Solving for vi:
vi^2 = 9.8
Taking the square root of both sides:
vi = √9.8
Calculating:
vi ≈ 3.13 m/s
Therefore, the initial velocity of the basketball player at the start of the leap is approximately 3.13 m/s.
To determine the initial velocity of the basketball player at the start of the leap, we can make use of the principles of physics and the equations governing the motion of objects.
The initial velocity (u) can be found using the formula for vertical motion:
v^2 = u^2 + 2as
where:
v = final velocity (which is zero at the highest point of the leap)
u = initial velocity
a = acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2)
s = displacement or height (0.5 m)
Since the final velocity (v) is zero at the highest point of the leap, the equation simplifies to:
0 = u^2 + 2as
Rearranging the equation to solve for u:
u^2 = -2as
Taking the square root of both sides:
u = √(-2as)
Substituting the given values:
u = √(2 * 9.8 * 0.5)
Calculating the values within the square root:
u = √9.8
Taking the square root:
u ≈ 3.13 m/s
Therefore, the initial velocity of the basketball player at the start of the leap is approximately 3.13 m/s.