During a practice match, a softball pitcher threw a ball. The height of the throw can be modelled by the equation $h = - 16 t^{2} + 24 t + 1$ , where h = height in feet and t = time in seconds. How long does it take for the ball to reach a height of 6 feet for the first time?

2.2, 3.8 secs

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5.4,6.2 secs

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0.25,1.25 secs

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7,5 secs

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Solution

The correct option is C
0.25,1.25 secs

Given height = 6
$\Rightarrow - 16 t^{2} + 24 t + 1 = 6$

$\Rightarrow 16 t^{2} - 24 t + 5 = 0$

$\Rightarrow (4 t - 1) (4 t - 5) = 0$

$t = \frac{1}{4}, \frac{5}{4} o r 0.25, 1.25$

It takes 0.25 secs to first reach a height of 6 feet. After 1.25 secs of being thrown, the ball again reaches a height of 6 feet from the ground because of free fall.

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Q. During a practice match, a softball pitcher throws a ball whose height can be modelled by the equation

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During a practice match, a softball pitcher threw a ball. The height of the throw can be modelled by the equation h=−16t2+24t+1, where h = height in feet and t = time in seconds. How long does it take for the ball to reach a height of 6 feet for the first time?

During a practice match, a softball pitcher threw a ball. The height of the throw can be modelled by the equation $h = - 16 t^{2} + 24 t + 1$ , where h = height in feet and t = time in seconds. How long does it take for the ball to reach a height of 6 feet for the first time?