Answer to Question #254673 in Mechanics | Relativity for gold

Question

2. A place-kicker must kick a football from a point 36.0 m from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0° to the horizontal. A. By how much does the ball clear or fall short of clearing the crossbar? B. Does the ball approach the crossbar while still rising or while falling?

Answer 1

A. From the particle under constant velocity model in the x direction, find the time at which the ball arrives at the goal:

$x_f = x_i + v_{xi}t \\ t = \frac{x_f-x_i}{x_{xi}} \\ = \frac{36.0-0}{20 \times cos(53°)} \\ =2.99 \;s$

From the particle under constant acceleration model in the y direction, find the height of the ball at this time

$y_f =y_i +v_{yi}t + \frac{1}{2}a_yt^2 \\ y_f = 0 + 20 \times sin(53°) \times 2.99 - \frac{1}{2} \times 9.80 \times 2.99 \\ = 3.94 \;m$

Therefore, the ball clears the crossbar by

$3.94 -3.05=0.89 \;m$

Use the particle under constant acceleration model to find the time at which the ball is at its highest point in its trajectory

$v_{vf} =v_{yi} -gt \\ t = \frac{v_{yi}-v_{yf}}{g} \\ = \frac{20.0 \times sin(53.0°) -0}{9.80} \\ = 1.63 \;s$

Because this is earlier than the time at which the ball reaches the goal, the ball clears the goal on its way down.