A high jumper goes into her jump at 2.37 m/s at an angle of 56.6° to the ground. How long is she in the air in seconds

Guest Nov 6, 2018

#1**+2 **

You can use the constant acceleration equation: v = u + a*t where u = initial vertical velocity = 2.37*sin(56.6°) m/s here; v = final vertical velocity = -u (when she hits the ground again); a = -9.8m/s^{2} (acceleration due to gravity); t = time in the air:

-2.37*sin(56.6°) = 2.37*sin(56.6°) - 9.8*t

t = 2*2.37*sin(56.6°)/9.8 s

Alan Nov 6, 2018

#1**+2 **

Best Answer

You can use the constant acceleration equation: v = u + a*t where u = initial vertical velocity = 2.37*sin(56.6°) m/s here; v = final vertical velocity = -u (when she hits the ground again); a = -9.8m/s^{2} (acceleration due to gravity); t = time in the air:

-2.37*sin(56.6°) = 2.37*sin(56.6°) - 9.8*t

t = 2*2.37*sin(56.6°)/9.8 s

Alan Nov 6, 2018