1. The problem statement, all variables and given/known data 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!