Question 312199: A model rocket is shot into the air and its path is approximated by , where h is the height of the rocket above the ground in metres and t is the elapsed time in seconds.
a. When will the rocket hit the ground?
b. What is the maximum height of the rocket?
Found 2 solutions by nyc_function, solver91311:
Answer by nyc_function(2741)   (Show Source):
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Can't do this for you except in a very general sense. You didn't provide the height function nor did you provide the initial velocity and initial height at launch from which to develop the specific function for the situation described.
However, in general, the height of a projectile launched vertically from near the surface of the earth is given by:
\ =\ -4.9t^2\ +\ v_ot\ +\ h_o)
where ) is the height in metres at time  ,  is the initial velocity at time  expressed in metres per second, and  is the initial height in metres at time
Part a:
The ground is  , so to determine when the rocket will hit the ground, solve:
Note: This quadratic has two roots. If  , which is to say the rocket was launched from the ground, then one of the roots will be zero and will correspond to the time of launch. The other root is a positive number of seconds corresponding to the time it took for the rocket to ascend to it's maximum height and return to earth. If  , then the rocket was launched from some sort of platform some finite distance above the ground. The equation will have two roots, one of which will be negative can be discarded because it corresponds to some time prior to the launch. The positive root will be the time to return to earth. If  , the rocket was launched from below ground level, perhaps from a silo or some such. In this case the equation will have two positive roots, the smaller of which corresponds to the elapsed time between launch and when the rocket emerged from the hole in the ground, and the larger of which corresponds to the return to earth.
Part b:
The quadratic equation represents a parabola that is concave downward (negative lead coefficient), therefore the vertex of the parabola represents a maximum point.
There are two ways to determine the maximum point.
If you are familiar with the calculus, you can take the first derivative of the function and set it equal to zero:
\ =\ -9.8t\ +\ v_o)
And then evaluate the function at that value:
\ =\ -4.9\left(\frac{v_o}{9.8}\right)^2\ +\ v_o\left(\frac{v_o}{9.8}\right)\ +\ h_o)
Or, if haven't been exposed to the concept of the derivative, use the formula for the vertex of a parabola in standard form, \ =\ ax^2\ +\ bx\ +\ c) , namely  :
and again, evaluate the function at the derived value:
John
|
|
|
