How far will the player be if she continues to run after the ball

[dani123]

Homework Statement

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?

Homework Equations

d_v=1/2*at²
d_h=V_h*Δt
Kinetic equation d=V_i*t+ 1/2*at²
d_h=-V²*sin2θ /g
sinθ=opp/hyp
cosθ=adj/hyp
Δt=-2Vsinθ / g

The Attempt at a Solution

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

d_h=cos(50)*22=14.1m
d_v=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!

 

[anathema]

I can confirm that the equations used are appropriate for solving this problem. The significant figures also appear to be respected, as the final answer is rounded to two decimal places, which matches the precision of the given values. Great job!