In American football, the ball is punted by dropping and kicking it before it hits the ground. The height ht of the football above the ground in meters t seconds after being punted is affected by gravity and by the punter's kick, and can be represented as the difference of two functions: at=4.9t2, which specifies the effect of gravity on the height of the ball, and vt=13.7t+0.5, which specifies the effect of the punt on the ball. With these two functions, ht=vt-at, how high above the ground was the ball when it was punted?

The ball was 0.5 m above the ground before it was punted.

Step-by-step explanation:

Since at = 4.9t² specifies the effect of gravity on the height of the ball, and vt = 13.7t + 0.5 specifies the effect of the punt on the ball, the height ht of the ball above the ground after the punt is  ht=vt-at

ht = 13.7t + 0.5 - 4.9t². To find the height ht above the ground at which the ball was punted, we insert t = 0 which is the start time of the punt into the equation for ht. So with t = 0, ht becomes

ht = 13.7t + 0.5 - 4.9t²

ht = 13.7 × 0 + 0.5 - 4.9 × 0²

ht = 0 + 0.5 - 0

ht = 0.5 m

So the ball was 0.5 m above the ground before it was punted.

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answer: sine

step-by-step explanation: