Question

The punter stands on the 22 yard (roughly a meter) line to the left of the 50 yard line and the ball is kicked with an initial velocity of 25.3 m/s at an angle of 58.5° above the ground. Where does the ball land?

Answer 1

The ball will land at approximately the `14` yard line on the opposite side of the field.

Explanation:

launch angle: `58.5^"o"`
initial velocity: `v_i=25.3"m/s"`

Vertical Motion Required to Find Time

initial vertical velocity: `v_"iy"=25.3"m/s"*sin58.5^"o"=21.6"m/s"`

final vertical velocity: `-21.6"m/s"`

time:
`v_"fy"=v_"iy" + "gt"`

`-21.6"m/s"=21.6"m/s"+(-9.81"m/s"^2)t`

`-43.2"m/s"=(-9.81"m/s"^2)t`

`(-43.2"m/s")/(-9.8"m/s"^2)=t`

`t=4.40"s"`

Horizontal Motion

initial horizontal velocity: `v_"ix"=25.3"m/s"*cos58.5^"o"=13.2"m/s"`

range: `x=x_"ix"*t=13.2"m/s"*4.40"s"=58.1"m"`

Because an American football field is marked in yards, convert `58.1"m"` to yards.

`58.1"m"xx(1.094"yd")/(1"m")=63.6~~64"yards"`

The ball started out on the `22` yard line and traveled an additional `64` yards. An American football field is `100` yards in length. The `50` yard line is in the center.

The location of the ball at the end of its flight will be `(50-22)-(50-36)=14`

The ball will land at approximately the `14` yard line on the other side of the `50` yard line.