Problem 4-21-Projectile - Part 8 - Solution

A football is placed on a line \(25\; m\) from the goalpost. The placement kicker kicks the ball directly toward the goalpost, giving it a velocity of \(21.0\; m/s\) at an angle of \(47.0^\circ\) above the horizontal. The horizontal bar of the goalpost is \(3.0\; m\) above the field. How far above or below the horizontal bar of the goalpost will the ball travel?

Diagram of football being kicked and all forces acting on it.

Accumulated Solution

\(v_{0_x}= 21 \cos 47 = (21) \cos47 = 14.32 \;m/s \\ v_{0_y}= 21 \sin 47 = (21) \sin47 = 15.36 \;m/s \\ x = v_{x^t} \\\; t = x/v_x = 25 \;m/14.32 \;m/s = 1.75\; s\)

\(v = v_0 + at \\ s = v_{0^t} + (1/2)at^2 \\ v^2 = v_{0^2} +2as\)

\(y = v_{0^t} + (1/2)at^2 = (15.36\; m/s)(1.75\; s) + (1/2)(-9.8 m/s^2)(1.75)^2 = 11.9 \;m\)

 

Distance above the crossbars \(= 11.9 \;m - 3.0\; m = 8.9 \;m\)

 

You have completed this problem.