A Standing Vertical Jump Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet left the floor.) Griffith weighed 890 N (200 Ib).
(a) What is his speed as he leaves the floor?
(b) If the time of the part of the jump before his feet left the floor was 0.300 s, what was his average acceleration (magnitude and direction) while he was pushing against the floor?
(c) Draw his free-body diagram (see Section 4.6). In terms of the forces on the diagram, what is the net force on him? Use Newton’s laws and the results of part
(b) To calculate the average force he applied to the ground.
The correct answer and explanation is :
We will solve the problem step by step using kinematic equations and Newton’s laws.
(a) Speed as he leaves the floor
Using the kinematic equation for motion:
[
v^2 = u^2 + 2as
]
where:
- ( v ) = final velocity (velocity at takeoff)
- ( u = 0 ) (initial velocity before the jump)
- ( a = -9.81 \, \text{m/s}^2 ) (acceleration due to gravity, acting downward)
- ( s = 1.2 \, \text{m} ) (vertical displacement)
Rearrange to solve for ( v ):
[
v = \sqrt{2gs} = \sqrt{2(9.81)(1.2)}
]
[
v = \sqrt{23.544} = 4.85 \text{ m/s}
]
So, his speed as he leaves the floor is 4.85 m/s.
(b) Average acceleration while pushing against the floor
Using the kinematic equation:
[
v = u + at
]
where:
- ( v = 4.85 \, \text{m/s} ) (from part a)
- ( u = 0 )
- ( t = 0.300 \, \text{s} )
Solving for ( a ):
[
a = \frac{v – u}{t} = \frac{4.85 – 0}{0.300}
]
[
a = 16.17 \, \text{m/s}^2
]
Thus, the average acceleration while pushing against the ground is 16.17 m/s² upward.
(c) Free-Body Diagram and Average Force
During the push-off phase, Griffith experiences:
- His weight, ( W = mg = 890 ) N (downward)
- The normal force exerted by the ground, ( F_N ) (upward)
Using Newton’s Second Law:
[
F_{\text{net}} = ma
]
[
F_N – W = ma
]
Solving for ( F_N ):
[
F_N = W + ma
]
Substituting known values:
[
F_N = 890 + \left(\frac{890}{9.81} \times 16.17\right)
]
[
F_N = 890 + (90.72 \times 16.17)
]
[
F_N = 890 + 1467.9 = 2357.9 \text{ N}
]
So, the average force he applied to the ground is 2358 N (upward).
Explanation (300 Words)
Darrell Griffith’s standing vertical jump is analyzed using kinematic equations and Newton’s laws. First, to determine his takeoff speed, we apply the equation of motion ( v^2 = u^2 + 2as ). Given that the vertical displacement is 1.2 m and acceleration due to gravity is ( 9.81 ) m/s² downward, solving for ( v ) gives 4.85 m/s. This is his speed as he leaves the floor.
Next, we determine the average acceleration while pushing against the floor. Using ( v = u + at ) with an initial velocity of 0 and a push-off time of 0.300 s, we find that his acceleration is 16.17 m/s² upward.
For force analysis, we consider his weight, which is 890 N, and the normal force exerted by the ground. According to Newton’s Second Law, the net force on Griffith is ( ma ), where ( m = \frac{W}{g} = \frac{890}{9.81} \approx 90.72 ) kg. The force required to accelerate him at 16.17 m/s² is calculated as ( F = ma ), resulting in an additional 1467.9 N. Adding his weight, the total force applied to the ground is 2358 N.
The free-body diagram consists of his weight acting downward and the normal force from the ground acting upward. The net force during the jump’s preparation phase leads to his acceleration. This analysis illustrates how forces and kinematics determine his incredible vertical jump performance.