A coin is dropped from a height of 750 feet. The height, s (ft), at time t (sec), is given by $s = - 16 t^{2} + 750$ $s=-16t^2+750$ . How long does it take for the coin to hit the ground?

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A coin is dropped from a height of 750 feet. The height, s (ft), at time t (sec), is given by $s = - 16 t^{2} + 750$ $s=-16t^2+750$ . How long does it take for the coin to hit the ground?

& what is the velocity when the coin hits the ground?

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Answer 1 · 2016-11-01T23:41:53

$6.85$ $6.85$ sec

Explanation:

At ground level, $s = 0$ $s=0$

$s = 0 \Rightarrow - 16 t^{2} + 750 = 0$ $s=0 => -16t^2+750 = 0$
$:. 16t^2 = 750$
$:. t^2 = 750/16$
$:. t^2 = 375/8$
$:. t = +-sqrt(375/8)$
$:. t = +-6.85$ (2dp)

We need $t>0$ , and so $t=6.85$ s (2dp)

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A coin is dropped from a height of 750 feet. The height, s (ft), at time t (sec), is given by $s = - 16 t^{2} + 750$ $s=-16t^2+750$ . How long does it take for the coin to hit the ground?

& what is the velocity when the coin hits the ground?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

A coin is dropped from a height of 750 feet. The height, *s* (ft), at time *t* (sec), is given by s=-16t^2+750s=−16t2+750s=-16t^2+750. How long does it take for the coin to hit the ground?

& what is the velocity when the coin hits the ground?

1 Answer

Explanation:

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Impact of this question

A coin is dropped from a height of 750 feet. The height, s (ft), at time t (sec), is given by $s = - 16 t^{2} + 750$ $s=-16t^2+750$ . How long does it take for the coin to hit the ground?