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. How far from the punter is the football when it's at its max height?

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Answer 1 · 2015-09-25T16:36:09

$29.1 m$ $29.1"m"$

To find the time to reach maximum height use:

$v = u + a t$ $v=u+at$

The vertical component of velocity is $V sin θ$ $Vsintheta$ where $θ$ $theta$ is the angle to the ground.

$a = g = - 9.8 m/s/s$ $a=g=-9.8"m/s/s"$

So this becomes:

$0 = 25.3 sin (58.5) - 9.8 t$ $0=25.3sin(58.5)-9.8t$

$t = \frac{- 21.57}{- 9.8} = 2.2 s$ $t=(-21.57)/(-9.8)=2.2"s"$

The horizontal component of velocity is a constant $25.3 cos (58.5)$ $25.3cos(58.5)$ .

So distance = velocity x time:

$d = 25.3 cos (58.5) \times 2.2 = 29.1 m$ $d=25.3cos(58.5)xx2.2=29.1"m"$