Question

If Michael Jordan has a vertical leap of 1.29 m, then what is his take off speed and his hang time (total time to move upwards to the peak and then return to the ground)?

• A. 4m/s, 2.33 sec
• B. 5.03 m/s, 1.03 sec
• C. 5.03m/s, 0.78 sec
• D. 6.1m/s, 6.78 sec

Answer 1

The correct option is B

5.03 m/s, 1.03 sec

$v_f^2=v_i^2+2\times a\times d$

$(0m/s{)}^2=v_i^2+2\times (-9.8m/s^2)\times (1.29m)$

$0m^2/s^2=v_i^2-25.28m^2/s^2$

$25.28m^2/s^2=v_i^2$

$v_i=5.03m/s$

To find hang time, find the time to the peak and then double it.

$v_f=v_i+a\times t$

$0m/s=5.03m/s+(-9.8m/s^2)\times t_{up}$

$-5.03m/s=(-9.8m/s^2)\times t_{up}$

$\frac{(-5.03m/s)}{(-9.8m/s^2)}=t_{up}$

$t_{up}=0.513s$

Hang time $=1.03s$