How Does Projectile Motion Determine a Football's Field Goal Success?

[kevinlikesphysics]

5. [SFHS99 3.P.38.] A place kicker must kick a football from a point 36.0 m (about 40.0 yd) from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 25.0 m/s at an angle of 45° to the horizontal.

(a) By how much does the ball clear or fall short of the crossbar? (Give a positive answer if the ball clears the crossbar, or a negative answer if it falls short of the crossbar.) _______ meters

(b) Does the ball approach the crossbar while still rising or while falling? (Type 'rising' or 'falling', or 'neither' if it falls short of the crossbar.)

What is the vertical component of velocity of the ball at this time? (Assume the positive direction is upward.) _______ m/s

 

[rocketboy]

try separating the problem into x and y components. make a chart with the displacement, acceleration, and velocity along the x and y axis. set up delta time as well (it's neither x or y). Now you can use your kinematics equations for the x-axis and y axis...essentially it becomes two 2-D problems. Then you can solve for a variable using one, and sub that into the other. Make sure you draw an accurate diagram to help you set up the x and y components correctly.

REMEMBER:

acceleration along the x-axis is zero
acceleration along the y-axis is the acceleration due to gravity

the magnitudes of the velocity's for x and y components are found using your trig...this one is really easy since it's at 45 degrees.

Try it and post what you have done. Then if you are still stuck we can help point you back in the right direciton.

(I just learned this stuff myself last month so it's great practice for me!)

 

[quicknote]

(a) To solve this problem, we can use the equations of projectile motion. First, we need to find the time it takes for the ball to reach the crossbar. Using the equation t = 2v*sin(theta)/g, where v is the initial velocity and theta is the angle of launch, we can calculate that the time is 2.86 seconds.

Next, we can use the equation y = y0 + v0t + 1/2gt^2 to find the vertical displacement of the ball. y0 is the initial height, which is 0 in this case since the ball starts at ground level. v0 is the initial velocity in the y-direction, which is v*sin(theta). Plugging in the values, we get y = 0 + (25*sin(45))*2.86 + 1/2*(-9.8)*(2.86)^2 = 21.3 meters.

Since the crossbar is at a height of 3.05 meters, the ball clears the crossbar by 21.3 - 3.05 = 18.25 meters. Therefore, the answer is a positive 18.25 meters.

(b) The ball approaches the crossbar while falling, as we can see from the positive value of the vertical displacement.

The vertical component of velocity at this time can be found using the equation v = v0 + gt, where v0 is the initial velocity in the y-direction and g is the acceleration due to gravity. Plugging in the values, we get v = (25*sin(45)) + (-9.8)*(2.86) = 0 m/s. This means that at the time the ball reaches the crossbar, its vertical velocity is 0 m/s, indicating that it has reached its maximum height and is about to start falling.