If g = 9.81 m/s^2
then
.6 = (1/2)(9.81) t^2
t^2 = 0.1223
t = 0.35 seconds up
2 t = 0.70 seconds in the air
reason
F = m a
a = -g = -9.81
then
v = Vi - 9.81 t where v is speed up and Vi is initial speed leaving ground
then
H = Vi t - (9.81/2) t^2 where H is the amount up in tie t
at top v = 0
so
Vi = 9.81 t at the top
thenat top
H = 9.81 t^2 - (9.81/2) t^2
= (1/2) (9.81) t^2 like we said
My question is straight out of the textbook, and I'm struggling to understand it.
Surprisingly, very few athletes can jump more than 2 feet
(0.6 m) straight up. Use d = 1/2gt2 to solve for the time
one spends moving upward in a 0.6-m vertical jump.
Then double it for the “hang time”—the time one’s feet
are off the ground.
1 answer
1 answer