James asked in Science & MathematicsPhysics · 10 years ago

# Trajectory Problem! - Field Goal?

In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 3.048 meters above the ground, and the ball is kicked from ground level, 10.9728 meters horizontally from the bar.

If the ball is kicked at 37 degrees above the horizontal, what must its initial speed be if it is to just clear the bar? Express your answer in meters per second (m/s).

It may be useful to know that the previous part was this:

There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle?

The answer to that is 15.5 degrees above the horizontal.

I already know that the answer is not 43.5 m/s or 8.32 m/s.

Please be detailed; I know a lot of other physics students who will benefit from a clear example. Thank you in advance.

Relevance
• Anonymous
10 years ago

Voy = Vo sin 37

Vox = Vo cos 37

Using the kinematic equations:

Y = Yo + Voy t - (1/2)gt^2; Where Yo = 0

X = Xo + Vox t - (1/2)at^2; Where Xo = 0 and a= 0

Using the kinematic equation for x:

10.9728 = (Vo cos 37) t

Vo = (10.9728) / [(cos 37) t] <======= Vo as a function of t

Substituting this for Vo in the kinematic equation for Y

3.048 = [(10.9728)/ {t cos 37}](sin 37) t - (1/2)(9.8)t^2

3.048 = 10.9728( tan 37) - 4.9 t^2

Solve for t:

t^2 = 1.0654

t = 1.032196 <---- At this time, y = 3.048 and x = 10.9728

Substitute back into equation to solve for Vo

Vo = (10.9728) / [(cos 37) t]

Vo = 13.311 m/s

• 10 years ago

When u aim to just clear the bar, that would mean that the ball should have just attained its max height when it clears the bar. At this point, the ball according to the projectile motion would have covered half of the horizontal distance(range) that it would cover. So, basically u are looking to make the ball cover twice the distance between the bar and the place the ball is kicked from, i.e. 2 X 10.9728 m.

Below is the solution using eculator(Find Velocity from Angle, Distance in step 1 of the solve problem tool):

Using the expression(see reference):

v0 = (R/( sin(2θ)/g))1/2

where:

R is the Range = 2*10.9728 m

θ is the Launch Angle = 37 ° = 0.6457718232379019 rad

g is the Gravitational Acceleration = 9.8

we calculate the value for Velocity(v0):

v0 = (2*10.9728/(Sin[2*0.6457718232379019]/9.8))^(1/2)

∴ v0 = 14.957738931286203