SOLUTION: A projectile is fired vertically upward from a height of 600 feet above the ground. It's height h(t) in feet above the ground after t seconds is given by h(t)=-16t^2+803t+600 a

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Question 126761: A projectile is fired vertically upward from a height of 600 feet above the ground. It's height h(t) in feet above the ground after t seconds is given by h(t)=-16t^2+803t+600
a) Estimate when the height of the projectile is 5000 feet above the ground.
b) Determine when the projectile will be more than 5000 feet above the ground.
c) How long will the projectile be in flight?
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
h%28t%29=-16t%5E2%2B803t%2B600+

We want to know t when h(t) = 5000, so:

-16t%5E2%2B803t%2B600=5000
-16t%5E2%2B803t-4400=0

Use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-803+%2B-+sqrt%28+803%5E2-4%28-16%29%28-4400%29%29%29%2F%282%2A%28-16%29%29
x+=+%28-803+%2B-+sqrt%28363209%29%2F%28-32%29%29

There is no way to simplify this because 363209 has no repeated prime factors.

The calculator gives us approximations of 6.26 seconds and 43.93 seconds.

This means that the projectile will reach 5000 feet at 6.26 seconds, continue upward to its maximum height, then start back down again and reach 5000 feet on the way down at 43.93 seconds.

So the answer to part a is 6.26 seconds, and the answer to part b is from 6.26 seconds until 43.93 seconds.

Part c asks how long the projectile will be in flight. It started at 600 ft above the ground, so on the way back down it has to go past the 600 ft starting point and continue on to 0. In other words, solve h(t)=0 for t.


-16t%5E2%2B803t%2B600=0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-803+%2B-+sqrt%28+803%5E2-4%28-16%29%28600%29%29%29%2F%282%2A%28-16%29%29
x+=+%28-803+%2B-+sqrt%28683209%29%2F%28-32%29%29

The two results are -0.74 and 50.92. Quite obviously the negative result can be discarded because you can have an event that occurs before the experiment started, so the other result, 50.92 seconds is the correct value for part c.

Super-double-plus extra credit: At what time will the projectile reach maximum height? How high will that be?

Hint: The vertex of a parabola is at the point (-b%2F2a,f%28-b%2F2a%29).