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A projectile is fired straight up from ground level with an initial velocity of 112ft/s. It’s height, h, above the ground after t seconds is given by h=-16t2+112t. What is the interval of time during which the projectile’s height exceeds 192 feet?

User MadBoy
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2 Answers

3 votes

Answer:

A. 3 < t < 4

Explanation:

For edge:)

User Cmirian
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8.2k points
6 votes

Answer:

t = 3 s and t = 4 s

Explanation:

Given that,

A projectile is fired straight up from ground level with an initial velocity of 112ft/s. Its height as a function of time t is given by :


h=-16t^2+112t

We need to find the interval of time during which the projectile’s height exceeds 192 feet. It means put h = 192 feet

So,


-16t^2+112t=192\\\\-16t^2+112t-192=0

The above is a quadratic equation, it can be solved using middle term splitting as follow :


16 (t^2 - 7t + 12) = 0\\\\16 (t - 4)(t - 3) = 0\\\\t=4\ s\ \text{and}\ t=3\ s

Hence, at t = 3 s and t = 4 s, the projectile’s height exceeds 192 feet above the ground.

User Arnold Brown
by
8.9k points
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