I had this on a test, and I\'m trying to verify I knew how tosolveit. Would some

Question

I had this on a test, and I'm trying to verify I knew how tosolveit. Would someone please solve so I can check my answeragainsttheirs? I got an initial velocity of apple being 30.35 m/sand timethrown at 2.09

An arrow is shot at an angle 50 degrees from the horizontal withamagnitude of 45m/s. Someone is standing 150m downrange holdinganapple.

What initial velocity must the apple be thrown so that its vertexisat the same height of the arrow at 150m?

At what time must the apple be thrown to intercept the arrow?

Explanation / Answer

first is to find the time it takes the arrow to reach the person horizontal distance=150m horizontal speed=45*cos50=28.925 so time it takes=150/28.925=5.186sec now, we need to find the height of the arrow using motion equation: x=x+vit+1/2at^2 x=0+(45sin50)(5.186)-1/2*9.81*5.186^2=46.85m now, this height must match the maximum heigh of the apple max height= (v^2sin^2Ø)/2g=the angle is 90(going up) 46.85=v^2/(2*9.81) 1.545m/s

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