56.7k views
3 votes
A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+218x+75

User SpellTheif
by
8.3k points

2 Answers

4 votes

Answer i got it wrong it was 5.94

Explanation:

User Ramesh Ponnusamy
by
8.5k points
5 votes

Answer:

6.81s

Explanation:

First you must know that the velocity of the body is zero at the maximum height.

v =dy/dx

v= -32x+218

Since v = 0

-32x+218 =0

32x =218

x= 218/32

x = 6.81s

Hence the rocket reaches its maximum height after 6.81s

User TestTester
by
7.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories