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anastassius [24]
3 years ago
12

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 w

here t is time (in seconds).
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
3 0

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

t = 5.0625

Step-by-step explanation:

Given

h(t)=-16t^2+80t+5

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

0=-16t^2+80t+5

Reorder

-16t^2+80t+5 = 0

Using quadratic formula, we have:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Where

a = -16      b =80      c = 5

So, we have:

t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}

t = \frac{-80\±\sqrt{6400 +320}}{-32}

t = \frac{-80\±\sqrt{6720}}{-32}

t = \frac{-80\±82.0}{-32}

This gives:

t = \frac{-80+82.0}{-32} or t = \frac{-80-82.0}{-32}

t = \frac{2}{-32} or t = \frac{-162}{-32}

t = -\frac{2}{32} or t = \frac{162}{32}

But time can not be negative.

So, we have:

t = \frac{162}{32}

t = 5.0625

<em>Hence, time to hit the ground is 5.0625 seconds</em>

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