# Math Help - Projectile Motion #1

1. ## Projectile Motion #1

Can someone help me with this problem. Thanks a lot!
Side is going to attempt a field goal by kicking a football from a point 40 yards out from the goal. The ball must clear the crossbar that is 3m off of the ground. Skid kicks the ball with a velocity of 20m/s at an angle of 45 degrees.
a)Does the ball clear the crossbar?
b)By how much does it clear or fall short of the crossbar?
c)Is the ball rising or falling as it approaches the crossbar?

What I did so far: y = ynot + vnot t +1/2 at^2
I then solve for t = squaroot 2(y-ynot)/9.8 and got .78. Then I took 20 x .78 and got 15.62 m but this is wrong. Can you tell me what I did wrong or how to work this problem out. Thanks again!!

2. Originally Posted by jsu03
Can someone help me with this problem. Thanks a lot!
Side is going to attempt a field goal by kicking a football from a point 40 yards out from the goal. The ball must clear the crossbar that is 3m off of the ground. Skid kicks the ball with a velocity of 20m/s at an angle of 45 degrees.
a)Does the ball clear the crossbar?
b)By how much does it clear or fall short of the crossbar?
c)Is the ball rising or falling as it approaches the crossbar?

What I did so far: y = ynot + vnot t +1/2 at^2
I then solve for t = squaroot 2(y-ynot)/9.8 and got .78. Then I took 20 x .78 and got 15.62 m but this is wrong. Can you tell me what I did wrong or how to work this problem out. Thanks again!!

first solve for t ...

$t = \frac{\Delta x}{v_x}$

$t = \frac{40}{20\cos(45)}$

use that value of t to find y ...

$y = y_0 + v_{0y} \cdot t - \frac{1}{2}gt^2$

$y = 20\sin(45) \cdot t - 4.9t^2$

if $y > 3$ , it clears the crossbar ... if not, oh well, they lose.

to answer the last question, find $v_f$ in the y-direction at the time you calculated above.

3. Hey, I did exactly what u told me and got 2.8 s, then I plug that into the equation and got 1.18, but the correct answer was .54m so can u tell me what I did wrong again.
40/20cos45= 2.8s
20sin45(2.8)-4.9(2.8)^2=1.18.....
for c) V= vnot +at
20- 9.8(2.8)
=-7.44 ?

4. Originally Posted by jsu03
Hey, I did exactly what u told me and got 2.8 s, then I plug that into the equation and got 1.18, but the correct answer was .54m so can u tell me what I did wrong again.
40/20cos45= 2.8s
20sin45(2.8)-4.9(2.8)^2=1.18.....
for c) V= vnot +at
20- 9.8(2.8)
=-7.44 ?
ok ... forgot to change yards to meters

40 yds = 36.576 m

so $t = 2.59$ sec

$y = 20\sin(45) \cdot t - 4.9t^2 = 3.8$ m