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serious [3.7K]
3 years ago
10

If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per​ second, then its height h aft

er t seconds is given by the equation h equals negative 16 t squared plus 112 t plus 128. After how many seconds does the object hit the​ ground? Round to the nearest tenth of a second.
Physics
1 answer:
Marina CMI [18]3 years ago
8 0

Explanation:

The equation of motion of an object is given by :

h(t)=-16t^2+112t+128

Where

t is the time in seconds

We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0

So,

-16t^2+112t+128=0

-t^2+7t+8=0

On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.

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Answer:

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Explanation:

To solve the problem, we can use the following SUVAT equation:

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where

v is the final velocity

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Answer:

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