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dani123
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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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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!
Report Post