1. The problem statement, all variables and given/known data(adsbygoogle = window.adsbygoogle || []).push({});

A soccer player running at 6.8 m/s kicks a soccer ball straight out in front of her so that it travels with an initial velocity of 22m/s at an angle of 50° with respect to the horizontal. If the soccer player continues to run at 6.68 m/s in the same direction that the ball was kicked, how far will she be from the ball when it strikes the ground?

2. Relevant equations

dv=1/2*at2

dh=Vh*Δt

Kinetic equation d=Vi*t+ 1/2*at2

dh=-V2*sin2θ /g

sinθ=opp/hyp

cosθ=adj/hyp

Δt=-2Vsinθ / g

3. The attempt at a solution

First, I started by finding the length of time the ball was in the air before hitting the ground.

dh=cos(50)*22=14.1m

dv=sin(50)*22=16.9m

Δt=14.1m/22m/s=0.64s

Then I found the distance traveled by the ball

d= (22m/s)*(0.64s)+ 1/2 (9.8)*(0.64s)2= 16.1m

and then I calculated the distance traveled by the player

d=(6.68m/s)*(0.64s)=4.28m I used this equation because there was no vertical distance to be measure for the player.

Then I found that the player would be d=16.1m-4.28m= 11.8m from the ball

I would like for someone to verify that the proper equations were used and that the significant figures are being respected. Thanks so much in advance!

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# Homework Help: Distance between player and ball- projectile motion

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