Answer to Question #140661 in Physics for Celeste
2020-10-26T14:49:54-04:00
How do I find the average vertical velocity (m/s) of a projectile shooting a ball at 15m/s with a 60 degree angle. I am trying to find its average vertical velocity at 0.4s and 4.41m vertical position.
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2020-10-27T11:15:20-0400
v y = v sin 60 − g t v y ( 0.4 s ) = 15 sin 60 − ( 9.8 ) ( 0.4 ) = 9.07 m s v_y=v\sin{60}-gt\\v_y(0.4s)=15\sin{60}-(9.8)(0.4)=9.07\frac{m}{s} v y = v sin 60 − g t v y ( 0.4 s ) = 15 sin 60 − ( 9.8 ) ( 0.4 ) = 9.07 s m
h = v t sin 60 − 0.5 g t 2 4.41 = 15 sin 60 t − 0.5 ( 9.8 ) t 2 t = 0.4 s h=vt\sin{60}-0.5gt^2\\4.41=15\sin{60}t-0.5(9.8)t^2\\t=0.4\ s h = v t sin 60 − 0.5 g t 2 4.41 = 15 sin 60 t − 0.5 ( 9.8 ) t 2 t = 0.4 s
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