MATH SOLVE

2 months ago

Q:
# a disc is thrown into the air with an upward velocity of 62 ft/s.its height h in feet after t seconds is given by the function h=-16tsquared +20t+6. What is the maximum height the disc reaches? How long will it take the disc to reach the maximum height? How long does it take for the disc to be caught 3 feet off the ground?

Accepted Solution

A:

Given that the height of the disc is modeled by the function:

h=-16t^2+20t+6

a] The maximum height will be as follows:

At maximum height:

h'(t)=0

from h(t)

h'(t)=-32t+20=0

thus

t=20/32

t=5/8 sec

thus the maximum height will be:

h(5/8)=-16(5/8)^2+20(5/8)+6

=12.25 ft

b] How long will it take the disc to reach the maximum height?

time taken to reach maximum height will be:

from h(t)

h'(t)=-32t+20=0

thus

t=20/32

t=5/8 sec

thus time taken to reach maximum height is t=5/8 sec

c]How long does it take for the disc to be caught 3 feet off the ground?

h(t)=-16t^2+20t+6

but

h(t)=3

thus

3=-16t^2+20t+6

solving for t we get:

0=-16t^2+20t+3

factoring the above we get:

t=5/8-/+√37/8

t=-1.5256 or 2.776

since there is no negative time we pick t=2.776

Hence time taken for the disc to be caught 3 ft from the ground is 2.776 ft

h=-16t^2+20t+6

a] The maximum height will be as follows:

At maximum height:

h'(t)=0

from h(t)

h'(t)=-32t+20=0

thus

t=20/32

t=5/8 sec

thus the maximum height will be:

h(5/8)=-16(5/8)^2+20(5/8)+6

=12.25 ft

b] How long will it take the disc to reach the maximum height?

time taken to reach maximum height will be:

from h(t)

h'(t)=-32t+20=0

thus

t=20/32

t=5/8 sec

thus time taken to reach maximum height is t=5/8 sec

c]How long does it take for the disc to be caught 3 feet off the ground?

h(t)=-16t^2+20t+6

but

h(t)=3

thus

3=-16t^2+20t+6

solving for t we get:

0=-16t^2+20t+3

factoring the above we get:

t=5/8-/+√37/8

t=-1.5256 or 2.776

since there is no negative time we pick t=2.776

Hence time taken for the disc to be caught 3 ft from the ground is 2.776 ft