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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

Best Answer 

 #1
avatar+27139 
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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/s2 (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
avatar+27139 
+1
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/s2 (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

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