A toy rocket is launched from the top of a building 95 feet tall at an initial velocity of223 feet per second. ​a) Give the function that describes the height of the rocket in terms of time t. ​b) Determine the time at which the rocket reaches its maximum​ height, and the maximum height in feet. ​c) For what time interval will the rocket be more than 851 feet above ground​ level? ​d) After how many seconds will it hit the​ ground?

Question
Asked Oct 25, 2019
A toy rocket is launched from the top of a building 95 feet tall at an initial velocity of
223 feet per second.
 
​a) Give the function that describes the height of the rocket in terms of time t.
 
​b) Determine the time at which the rocket reaches its maximum​ height, and the maximum height in feet.
 
​c) For what time interval will the rocket be more than 851 feet above ground​ level?
 
​d) After how many seconds will it hit the​ ground?
check_circleExpert Solution
Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

Acceleration due to gravity = 32  feet per second^2

Initial velocity=v0=223 feet per second. 

Initial height =h0=95 feet

Height at t second is h(t)

Answer(a): h(t)=-16t^2+223t+95

a
h(t) +vh
2
32
2+223t+95
2
h(t)=
h(t)-16t 2+223t+95
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a h(t) +vh 2 32 2+223t+95 2 h(t)= h(t)-16t 2+223t+95

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Step 2

When the rocket reaches maximum height then t=-b/(2a) [using vertex formula] With a=-16 and b=223

At t=6.96875 we then find the height....

h(t)=-16t 2+223t+95
223
t=
2(-16)
t=6.96875
h(6.96875)=-16(6.96875)+223(6.96875)+95
h(6.96875)=872.015625
help_outline

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h(t)=-16t 2+223t+95 223 t= 2(-16) t=6.96875 h(6.96875)=-16(6.96875)+223(6.96875)+95 h(6.96875)=872.015625

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