During a game of catch, you tell your friend---who is standing about 15 feet awa

Question

During a game of catch, you tell your friend---who is standing about 15 feet away from you---to go long. After waiting 3 seconds you fling the ball as hard as you can. Your friend manages to dive and catch the ball right at the edge of the field about 40 yards away. You estimate that this throw took about 6.0 seconds from the time your friend started to run. Being the curious person you are, this raises a number of interesting questions.How fast did you throw the ball?How high up did the ball get?

Explanation / Answer

How fast did you throw the ball=V = ?

DIstance coverd by your friend = S1 = 40yards = 36.5m

And you was standing away from him at the distance =S2=15feet = 4.5m

Total distance coverd will be = S = S1+S2 = 4.5m + 36.5m = 41m

Time taken = t = 6sec

We know that

V = S / t = 41m / 6sec = 6.83m/s

B) How hight up did the ball get = H=?

Applying second equation of motion along the vertical component

2as = Vf2 - Vi2

-2gH = Vf2 - VI2

H =02 - 6.832 / -20 at maximum height final velocity = Vf =0m/s

H = 2.33m

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