Projectile Motion Problem: Kicking a soccer ball over a fence

[salqmander]

Homework Statement

A soccer player is practicing their kick on a field. Initially at rest, an 0.8 kilogram ball is kicked directly toward a fence from a distance 25 meters away, as shown above. The ball's velocity as it leaves the kicker's foot is 19 m/s at an angle of 30 degrees above the horizontal. The top of the fence field is 2.5 meters high. The kicker's foot is in contact with the ball for 0.05 seconds. While in flight, the ball doesn't hit any other object, and air resistance is negligible.

Determine whether the ball will hit the fence. If it will, how high up the fence will it hit? If not, how far above the fence will it reach?

Relevant Equations

d = vt + .5at^2

initial velocity y component is (cos30) * 20.

t = 25m / ((cos30) * 20)m/s = 1.45 seconds

d = vt + .5at^2

v= 20sin30

v= 10 , d= 10(1.45s) + .5(-9.8m/s^s)(1.45s)^2

d=4.2m

4.2-2.5 = +1.7m, so the ball will not hit the fence

I need confirmation please

 

[PeroK]

Looks good, except you use $20 m/s$ instead of $19 m/s$ given in the question?

 

[salqmander]

oh I didn't catch that! i'll fix it, thank you

 

[haruspex]

initial velocity y component is (cos30) * 20.

Have another go (and I'm not referring to whether it's 19 or 20).

 

[salqmander]

should I use the equation
xf = xi + (vx)i delta t
for time and then
yf = yi + vyi delta t- 1/2g t ^2

 

[salqmander]

Have another go (and I'm not referring to whether it's 19 or 20).

is this right?

xf = xi + (vx)i delta t, xi=0

t = xf / vxi

t = 25m / 19m/s, t = 1.3 seconds
vyi = vi sin theta, vyi = 9.5m/s

yf = yi + vyi delta t- 1/2g t ^2

= 0m + 9.5m/s(1.5) - 0.5(9.8m/s^2)(1.5)^2

yf = 3.2m

3.2-2.5 = +0.7m

 

[haruspex]

is this right?

xf = xi + (vx)i delta t, xi=0

t = xf / vxi

t = 25m / 19m/s, t = 1.3 seconds
vyi = vi sin theta, vyi = 9.5m/s

yf = yi + vyi delta t- 1/2g t ^2

= 0m + 9.5m/s(1.5) - 0.5(9.8m/s^2)(1.5)^2

yf = 3.2m

3.2-2.5 = +0.7m

In post #4 I quoted one of your equations. Why am I finding fault with it?

 

[salqmander]

oh its sin not cos

 

[kuruman]

t = xf / vxi

t = 25m / 19m/s, t = 1.3 seconds

Also, do you see a similar problem above?

 

[salqmander]

Also, do you see a similar problem above?

yes, i solved for horizontal position not vertical position

 

[kuruman]

yes, i solved for horizontal position not vertical position

That's not it. You solved for the time of flight. Do you see what's wrong with it?

 

[PeroK]

Homework Statement: A soccer player is practicing their kick on a field. Initially at rest, an 0.8 kilogram ball is kicked directly toward a fence from a distance 25 meters away, as shown above. The ball's velocity as it leaves the kicker's foot is 19 m/s at an angle of 30 degrees above the horizontal. The top of the fence field is 2.5 meters high. The kicker's foot is in contact with the ball for 0.05 seconds. While in flight, the ball doesn't hit any other object, and air resistance is negligible.

Determine whether the ball will hit the fence. If it will, how high up the fence will it hit? If not, how far above the fence will it reach?
Relevant Equations: d = vt + .5at^2

initial velocity y component is (cos30) * 20.

t = 25m / ((cos30) * 20)m/s = 1.45 seconds

d = vt + .5at^2

v= 20sin30

v= 10 , d= 10(1.45s) + .5(-9.8m/s^s)(1.45s)^2

d=4.2m

4.2-2.5 = +1.7m, so the ball will not hit the fence

I need confirmation please

This solution was correct, apart from the typo(?) of y component, rather than x component. It was subsequently used as the x-component. And, the use of the wrong initial speed. Subsequent attempts seem to have deteriorated somewhat!