Question

If Rahul 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)?

Answer 1

Answer:

Here v=0 , s=1.29m ,u=? ,a=-9.8 m/s2

Now,

v2=u2+2asv2=u2+2as

u= 5.03m/s.

Now time going Up,

v=u+at

or

t=-u/a = 5.03/9.8 =.513 s

Total hang time = 2 * .513 =1.03s