During a practice match, a softball pitcher throws a ball whose height 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?

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 and 1.25secs, ball to go up and down respectively.

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Similar questions

Q. During a practice match, a softball pitcher throws a ball whose height 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?

Q. During a practice match, a softball pitcher throws a ball whose height 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?

During a practice match, a softball pitcher throws a ball, whose height can be modeled 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?

During a practice match, a softball pitcher throws a ball whose height can be modeled by the equation h = $- 16 t^{2}$ + 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?

Q. During a practice match, a softball pitcher throws a ball whose height can be modeled 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?

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During a practice match, a softball pitcher throws a ball whose height 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?

The correct option is C 0.25,1.25 secs Given height = 6 ⇒−16t2+24t+1=6 ⇒16t2−24t+5=0 ⇒(4t−1)(4t−5)=0 t=14,54 or 0.25, 1.25 It takes 0.25 secs and 1.25secs, ball to go up and down respectively.

During a practice match, a softball pitcher throws a ball whose height 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?

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 and 1.25secs, ball to go up and down respectively.