1. The problem statement, all variables and given/known data Jaime is outside playing with her dog, Skip. She throws a tennis ball at an angle of 60° with an initial speed of 10.2m/s and from a height of 196cm. If Skip was standing beside her when she threw the ball, how far would Skip have to run in order to catch the ball just the moment before it hits 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 dv= sin(60)*10.2m/s= 8.83m The actual vertical height for the dog is 1.96m+8.83m=10.8m t=√[10.8m/(0.5)*(9.8)]=1.48s Then dv=1/2*at2= 25.9m If someone could just verify my work and make sure my answers are correct that would be greatly appreciated. Also, if you could check my significant figures are being respected. Thanks so much in advance!