Trouble Setting Up Projectile Motion

[quality101]

Problem: A football is kicked from 36.0 m away from goal post at an angle of 53.0 deg. above horizontal at a speed of 20.0 m/s. The crossbar is at a height of 3.05 m high. a.) By how far does the ball clear or fall short of the crossbar? b.) Does the ball approach the crossbar while still rising or while falling?

So far, I have drawn my sketch of the problem and I have chosen the x-y origin to be at the point the football is kicked, x being positive and height of crossbar (y) is positive.

I have also drawn out a vector right angle triangle to find the trigonometric relationship between Vnaught-x, Vnaught-y, and Vnaught. From the problem, Vnaught = 20.0 and theta = 53.0 deg. I now have solved Vnaught-x to be 12.04 and Vnaught-y to be 15.97. Is Vnaught-y the max height of the ball at a distance of 12.04 meters from the origin? If so, where do I go from here?

Thanks,
Steve

 

[Pyrrhus]

HUGE HINTS!
You might want to calculate the time at max height, so you can tell if the ball is falling or rising when it crosses the crosshair. Also consider evaluation Y when x= 36, so you can see how high will the ball be, and if it will be higher than 3.05 m.

 

[M98Ranger]

Hi Steve,

It seems like you have made a good start in setting up the problem. To answer your first question, Vnaught-y does represent the maximum height of the ball at a distance of 12.04 meters from the origin. To find the distance the ball clears or falls short of the crossbar, you will need to use the kinematic equations for projectile motion.

The first equation to use is the equation for the vertical displacement, which is y = y0 + v0y*t - 1/2*g*t^2. In this case, y0 (initial height) is 0 since the ball is kicked from ground level. v0y (initial vertical velocity) is 15.97 m/s and g (acceleration due to gravity) is -9.8 m/s^2. We are looking for the time (t) when the ball reaches a height of 3.05 m (height of the crossbar). So we can plug in these values and solve for t.

3.05 = 0 + (15.97)*t - 1/2*(-9.8)*t^2
3.05 = 15.97t + 4.9t^2
4.9t^2 + 15.97t - 3.05 = 0

Using the quadratic formula, we get two possible values for t: 0.15 s and -3.14 s. Since time cannot be negative, we can ignore the second solution. So the ball reaches a height of 3.05 m after 0.15 seconds. We can now use this time in the horizontal displacement equation, x = x0 + v0x*t, to find the horizontal distance the ball travels.

x = 0 + (12.04)*0.15
x = 1.81 meters

Since the ball was kicked from a distance of 36.0 m, we can see that it falls short of the crossbar by 36.0 - 1.81 = 34.19 meters.

To answer your second question, the ball approaches the crossbar while falling. This is because the vertical velocity decreases due to gravity, causing the ball to eventually fall back down to the ground.

I hope this helps you solve the problem. If you have any further questions, don't hesitate to ask. Good luck!