Distance between player and ball- projectile motion

[dani123]

1. 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?

2. Homework 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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[collinsmark]

1. 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?

2. Homework 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.

That's the right approach.

dh=cos(50)*22=14.1m
dv=sin(50)*22=16.9m

Δt=14.1m/22m/s=0.64s

...But that's not the right way to implement it.

You'll need to break up the ball's movements into the x and y components.

The component that involves the up/down (vertical) direction is the one that determines how long the ball is in the air. That's the same component that involves acceleration due to gravity. The ball is accelerating in the vertical direction (and only in the vertical direction).

Then I found the distance traveled by the ball
d= (22m/s)*(0.64s)+ 1/2 (9.8)*(0.64s)2= 16.1m

That's not quite right either. The horizontal distance traveled by the ball does not [directly] involve acceleration due to gravity.

The horizontal component of the ball is not accelerating. It is moving at a constant velocity. And once you know the horizontal component of the ball's velocity, together with the time the ball is in the air, you can calculate the ball's distance.

 

[laurent711]

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[dani123]

Thank you so much for your help! I have used an example that I found online as a guide as well and hoping that you would be able to verify if I got the right idea.

I broke up the problem and started by focusing on the information we have from the ball:
V_ix=22m/s
θ=50°

R=V_i²sin2θ/g= 48.64m

then i went on to find the time of travel of the ball:
Δt=2|V_i|sinθ/g= 3.44s

Then I continued to look at the players distance travelled:
d_h=(3.44s)*(6.68m/s)= 22.98m

So I figured the player would be 48.64m-22.98m=25.66m away from the ball once it hit the ground.

Would this be correct? Thanks again for your time and help, it is greatly appreciated!

 

[Shivam123]

yes you have done it the right way,answer is correct.