Question

A model rocket is launched directly upward at a speed of 50 meters per second from a height of 8 meters. The function f(t)=−4.9t2+50t+8, models the relationship between the height of the rocket at different times, t, in seconds. When, in seconds after launch, will the rocket hit the ground? Round to two decimal places.

Answer 1

Answer:

$t=\frac{-50 -\sqrt{(50)^2 -4(-4.9)(8)}}{2*(-4.9)}= 10.36$

$t=\frac{-50 +\sqrt{(50)^2 -4(-4.9)(8)}}{2*(-4.9)}= -0.158$

And since the time can't be negative we have that:

$t= 10.36 seconds$

So the answer for this case would be after 10.36 seconds the rocket will hit the ground

Step-by-step explanation:

For this case we have the function of the height in terms of the time givenby:

$h(t) = -4.9t^2 +50 t+8$

And we want to known when the rockect hit the ground and for this case we need to solve the equation $h=0$

$-4.9t^2 +50 t+8 =0$

And as we can see we have a quadratic equation like this one:

$ax^2 +bx+c$

Where: $a =-4.9 , b =50 c =8$

And the general solutions are given by:

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

If we replace the values that we have we got:

$t=\frac{-50 -\sqrt{(50)^2 -4(-4.9)(8)}}{2*(-4.9)}= 10.36$

$t=\frac{-50 +\sqrt{(50)^2 -4(-4.9)(8)}}{2*(-4.9)}= -0.158$

And since the time can't be negative we have that:

$t= 10.36 seconds$

So the answer for this case would be after 10.36 seconds the rocket will hit the ground