Question

A punter can kick a football with an initial velocity of 48 meters per second. How many seconds will it take the ball to return to the ground? Use the formula h = vt - 16t^2

Answer 1

Using the equation of motion for a freely falling object and setting the height to zero, we find that it takes 3 seconds for the football to return to the ground after being kicked with an initial vertical velocity of 48 meters per second.

Step-by-step explanation:

To determine how many seconds it will take the football to return to the ground with an initial velocity of 48 meters per second, we need to use the equation of motion for a freely falling object: h = vt - 16t^2, where h is the height above the ground, v is the initial velocity in the vertical direction, and t is the time in seconds. Because we want to find the time when the ball returns to the ground, we set h = 0. This gives us the quadratic equation 0 = 48t - 16t^2.

Solving the Quadratic Equation

1. Divide every term by 16 to simplify the equation: 0 = 3t - t^2.
2. Rearrange the terms to get a standard quadratic form: t^2 - 3t = 0.
3. Factor the quadratic equation: t(t - 3) = 0.
4. Set each factor equal to zero and solve for t: t = 0 and t = 3.

The t = 0 solution corresponds to the initial time when the ball is kicked. The other solution, t = 3 seconds, is the time it takes for the ball to return to the ground. Hence, it will take 3 seconds for the ball to land back on the ground.

Answer 2

check the picture below.

`\bf \stackrel{h}{0}=\stackrel{v=48}{48t}-16t^2\implies 0=16t(3-t^2) \\\\[-0.35em] ~\dotfill\\\\ 0=16t\implies 0=t\impliedby \textit{when it first takes off from the ground} \\\\[-0.35em] ~\dotfill\\\\ 0=3-t^2\implies t^2=3\implies t=√(3)\impliedby \textit{when it came back down}`