A=3, C=8 Please explain steps in depth Path of a Rocket A model rocket is launch

Question

A=3, C=8
Please explain steps in depth Path of a Rocket A model rocket is launched with an initial velocity of 11[c] feet per second from the height of [a] feet. The unction s (t)11clt(a] models the path of the rocket where t represents the time, in seconds, after which the rocket is launched and s represents the height of the rocket at time t 1. Fill inthetable g of values for the function, and give a short interpretation for each point. 2. How long will it take for the rocket to return to the ground? 3. How long will it take the rocket to reach its maximum height? What is the maximum height reached by the rocket? Round all answers to one decimal place. Note: if [a-2 and [cJ-7, then s (t) =-16t2 + 117t + 2 Work this custom problem about the quadratic function using the techniques you learned in this module. MATH 111 Turn-In Assignment (7) Criteria Ratings Pts Initial submission demonstrates at least a rudimentary understanding of the solution method. 25 0 pts 18.0 pts100 pts Proficient Competent Novice 25.0 pts 25 0 pts 18 0 pts10 0 pts Proficient Competent Novice 25.0 pts Correct solution. Detailed explanation of work with the use of an equation editor for mathematical500pts 300 pts 20 0 pts writing. Use of complete sentences. APA format where needed Proficient Competent Novice Total Points: 100.0

Explanation / Answer

so you have given that a=3 and c=8 ,

so the equation becomes s(t) = -16t2+88t+3

now substitute the values give in table

for t= 0 , s(t) = -16*0 + 88*0 + 3 , whihc means s(0) = 3 ,, interpretation = this is the height at which roket was initailly launched

for t=2 , s(t) = -16*4 + 88*2 +3 whihc means s(2) = 115 , interpretation = the rocket has reached of 115 and gained 112 m height in 2 sec

for t=4 , s(t) = -16*16+ 88*4 +3 whihc means s(4) = 99 , here we can see that the rocket has hit the highest point and is returning back

for t=10 , s(t) = -16*100 +88*10 +3 whihc means s(10) = -717 , here we can say that rocket has reached the ground before 10 secs so its showing negavtive values which are not possible as once it hits the ground it stops

for s(t) =150 , 150=-16t2+88t+3 , so equation becomes 16t2-88t+147= 0 so this is in the form of ax2+bx+c = 0 where a = 16 , b =-88 , c=147 , use the formula to solve for x which is x=(-b+(squareroot(b2-4ac))) / 2a and x=(-b-(squareroot(b2-4ac))) / 2a , as x has no values for the above equation as square root of a negative number doesnot exist, so interpretation the rocket doesnot reach that height

for s(t)= 200 , same answer as above as sqaure root of a negative number doesnt exist , so rocket doesnt reach that height

2) for rocket to return to ground means , s(t) = 0 , so equation becomes 16t2-88t-3=0 , using the formula mentioned above , a=16, b =-88, c = -3 , so the value for t is 5.53 and -0.03 , as we know that negative values are not possible for time in real life , so the value is 5.53 sec

3) for obtaining the maximum height the derivative of the equation should be equal to 0 , so differentiate the equation by t whihc will result in s(t)=-32t+88 , whihc should be equal to 0 so 0=-32t+88 , so t= 88/32 =2.75 seconds

derivative of ax2 =2ax , and derivative of ax= a and derivative of a constant is 0 , so -16t2 becomes -32t and 88t becomes 88 and 3 becomes 0

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