Dennis the Menace is shooting a rock with his slingshot with an initial height of 5 feet. The height in feet, of the rock above the ground is given by s(t)=-16t^2+44t+5, where t is time in seconds and t is >= 0. At what time will the rock be 15 ft above the ground?

The rock will be 15ft above the ground 0.25 and 2.5secs after

The height of the rock above the ground is expressed according to the expression

In order to get the time the rock will be 15ft above the ground, we will substitute s(t) = 15 ft into the formula above;

Factorize the expression to get the value of "t"

2(4t−1)(2t−5) = 0

4t - 1 = 0

4t = 1

t = 0.25secs

Similarly 2t - 5 =0

2t = 5

t = 5/2 = 2.5secs

This shows that the rock will be 15ft above the ground 0.25 and 2.5secs after

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 32

The average velocity of the rock would be 32 ft per second.

layer a is the exosphere and layer b is the thermosphere.   is the answer i think but i hope this .

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