MATH SOLVE

5 months ago

Q:
# 13. A catapult launches a boulder wint an upward velocity of 122 feet per second the height of the boulder (h) in feet after t seconds is given by the function h(t)=-16t^2+122t+10. How long does it take the boulder to reach maximum height? What is the boulders maximum height? Round to the nearest hundredth if necessary.

Accepted Solution

A:

For this case we have the following expression:

h (t) = - 16t ^ 2 + 122t + 10

We look for the maximum of the function, for this, we derive:

h '(t) = - 32t + 122

We match zero:

-32t + 122 = 0

We cleared t:

t = 122/32

t = 3.8125 s

Then, the maximum height will be:

h (3.8125) = - 16 * (3.2185) ^ 2 + 122 * (3.2185) +10

h (3.8125) = 236.92 feet

Answer:

It takes the boulder to reach maximum height about:

t = 3.8125 s

the boulders maximum height is:

236.92 feet

h (t) = - 16t ^ 2 + 122t + 10

We look for the maximum of the function, for this, we derive:

h '(t) = - 32t + 122

We match zero:

-32t + 122 = 0

We cleared t:

t = 122/32

t = 3.8125 s

Then, the maximum height will be:

h (3.8125) = - 16 * (3.2185) ^ 2 + 122 * (3.2185) +10

h (3.8125) = 236.92 feet

Answer:

It takes the boulder to reach maximum height about:

t = 3.8125 s

the boulders maximum height is:

236.92 feet