If skier has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time?

Given:

a = -9.8 m/s^{2}

v_{f} = 0 m/s

d = 1.29 m

Find:

t = ??

(0 m/s)^{2} = v_{i}^{2} + 2*(-9.8 m/s^{2})*(1.29 m)

0 m^{2}/s^{2} = v_{i}^{2} - 25.28 m^{2}/s^{2}

25.28 m^{2}/s^{2} = v_{i}^{2}

v_{i} = 5.03 m/s

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

v_{f} = v_{i} + a*t

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

-5.03 m/s = (-9.8 m/s^{2})*t_{up}

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

t_{up} = 0.513 s

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up