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At t = 0, an object is projected with a speed v0 = 45 m/s at an angle q0 = 30.4° above the horizontal. The axes on the diagram show the x and y directions that are to be considered positive.
its horizontal velocity is 38.81m/s
at t = 10 s

2006-09-12 08:55:23 · 3 answers · asked by hardik p 1 in Science & Mathematics Physics

3 answers

vertical speed at t=0
v0=sin30.4*45=22m/s

v=v0+a*t=22-9.81*10=-75m/s

2006-09-12 09:06:28 · answer #1 · answered by Anonymous · 0 0

Hi. Try sin 30.4 times 45. This should be the initial vertical velocity and may guide you to your answer. (I usually don't answer, just hint. Sorry.)

2006-09-12 15:59:19 · answer #2 · answered by Cirric 7 · 0 0

The rate of upward or downward motion of air passing through a given pressure level.

2006-09-12 17:40:55 · answer #3 · answered by Anonymous · 0 1

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