SOLUTION: Ok so the problem is A rocket is launched from atop a 41 foot cliff with an initial velocity of 103 feet per second. The height of the rocket t seconds after launch is given by the
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: Ok so the problem is A rocket is launched from atop a 41 foot cliff with an initial velocity of 103 feet per second. The height of the rocket t seconds after launch is given by the
Log On
Question 981891: Ok so the problem is A rocket is launched from atop a 41 foot cliff with an initial velocity of 103 feet per second. The height of the rocket t seconds after launch is given by the equation . Graph the equation to find out how long after the rocket is launched it will hit the ground. Round to the nearest tenth of a second.
I know this is a large problem but please help and ill be very grateful.
I remember my math teacher asking this but it was 2 years ago so I forgot how to do it.
Please explain also. Found 2 solutions by rothauserc, Alan3354:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! The equation is
h = -16t^2 +103t + 41
this is a parabola that opens downward
to find time t when the rocket hits the round solve the following for t
-16t^2 +103t + 41 = 0
use quadratic formula,
t = (-b + or - square root(b^2 - 4ac)) / 2a
we get the following solutions for t
t = 6.813586792 or t = −0.376086792
we want the positive value for t, therefore
the rocket will hit the ground in 6.8 seconds
here is the graph
You can put this solution on YOUR website! Why is that called a rocket?
Rockets have thrust and accelerate upwards.
That's just a projectile, like a rock. Not a rocket.
