Question

A basketball star covers 3.05 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.95 m above the floor and is at elevation 0.890 m when he touches down again. What is his time of flight?

Answer 1

anonymous  4 years agoAny time you are mixing distance and acceleration a good equation to use is ΔY=Viyt+1/2at2 I would split this into two segments - the rise and the fall. For the fall, Vi = 0 since the player is at the peak of his arc and delta-Y is from 1.95 to 0.890. For the upward part of the motion the initial velocity is unknown and the final velocity is zero, but motion is symetrical - it takes the same amount of time to go up as it does to go down. Physiscists often use the trick "I'm going to solve a different problem, that I know will give me the same answer as the one I was actually asked.) So for the first half you could also use Vi = 0 and a downward delta-Y to solve for the time. Add the two times together for the total. The alternative is to calculate the initial and final velocity so that you have more information to work with.

Answer 2

Answer:

T = 0.90 s

Explanation:

Initially the position of center of mass is at 1.02 m

Then he reached to height 1.95 m

so total displacement of the player in vertical direction is

$y = 1.95 - 1.02 = 0.93 m$

now we know that it will reach to this displacement when final velocity becomes zero

$v_f^2 - v_i^2 = 2a y$

$0 - v_i^2 = 2(-9.81)(0.93)$

$v_i = 4.27 m/s$

now we have

$v_f - v_i = at$

$t_1 = \frac{0 - 4.27}{-9.81}$

$t_1 = 0.435 s$

Now when it will reach the ground again then his displacement is given as

$y = 1.95 - 0.890 = 1.06 m$

$\delta y = \frac{1}{2}gt^2$

$1.06 = \frac{1}{2}(9.81)t^2$

$t_2 = 0.465 s$

now total time of his motion is given as

$T = t_1 + t_2$

$T = 0.465 + 0.435$

$T = 0.90 s$