Question

A punter in a football game kicks a ball from the goal line at 60.0° from the horizontal at 25.0 m/s a)what is the hang time of the punt in the air? (4.41 s)b)How far down field does the ball land? (55.2 m)

Answer 1

ANSWER

a)4.41 s

b) 55.2 m

EXPLANATION

Part a

Velocity has the equation:

$v=u+at$

This is a two-dimentional motion problem, so we have to analyze each component of the velocity separately. Since we don't know the horizontal acceleration, we can just work with the vertical component to find the time. Let's draw a diagram to understand this situation better:

We know the initial velocity and the vertical acceleration. We could first find the time the punt is in the air until it reaches its maximum height, because at this point the vertical velocity is zero - this is because the object stops and starts speeding up in the opposite direction.

Therefore, for the vertical component of the velocity we have:

$\begin{gathered} v_y=u_y-gt \\ \text{ since }v_y=0 \\ 0=u_y-gt \end{gathered}$

Solving for t:

$t=\frac{u_y}{g}$

Now, how do we find uy? We just have to use the triangle formed by the velocity vector and its components. Note that uy is the opposite side to the given angle, so we use the sine:

$\begin{gathered} \sin 60\degree=\frac{u_y}{u} \\ u_y=u\sin 60\degree \end{gathered}$

So the time for the punt to reach the maximum height is:

$t=\frac{u\sin60\degree}{g}=\frac{25\cdot\sin 60\degree}{9.8}=2.21s$

This, however, is just half the time the punt is in the air. By inference, the time it takes for the punt to reach the ground will be the same and therefore, the total time it is in the air is:

$t=2\cdot2.21=4.42$

The difference in the decimals is because it depends on how you perform the calculations and how the partial results are rounded.

Part b

The horizontal velocity is constant, only the vertical component changes. So the horizontal velocity is:

$v_x=\frac{d}{t}$

We have the time, found in part a, and the horizontal velocity, because it is constant so it is the horizontal component of the initial velocity:

$v_x=u_x=u\cdot\cos 60\degree_{}$

So the horizontal distance the ball travels is:

$\begin{gathered} d=v_x\cdot t \\ d=u\cdot\cos 60\degree\cdot t \\ d=25\cos 60\degree\cdot4.42 \\ d=55.25m \end{gathered}$