college question

Question

A toy rocket is launched straight up from the top of a building 50 ft tall at an initial velocity of 200 ft per sec. give the function that describes the height of the rocket in terms of time t. Determine the time at which the rocket reaches its maximum height and the maximum height in feet. For what time interval will the rocket be more than 300 ft about the ground level? after how many seconds will it hit the ground

Answer 1

The general equation for initial velocity quadratics is:

 

h(t) = at^2 + v_ot + h_o

 

a = acceleration. This value is -16 if our units are feet or -4.9 in meters

v_o= initial velocity

h_o= initial height

 

so

 

h(t) = -16t^2 + 200t + 50

 

To find the maximum height, we must put the equation in vertex form:

 

h(t) = -16(t^2 - 12.5t) + 50

h(t) = -16(t^2 - 12.5t + 39.0625) + 50 -(-16•39.0625)

h(t) = -16(t - 6.25)^2 + 675

 

Therefore the vertex of this downward facing parabola is at (6.25, 675) so the max height of 675 feet occurs after 6.25 seconds

 

To find when the rocket reaches 300 feet, you could plot 

y1 = -16(t - 6.25)^2 + 675

and

y2 = 300

And find where the two graphs intersect (which should happen twice)

 

You could use a similar method to find when the graph hits the ground with y2 = 0 or alternatively, you could simply find the x-intercepts of the graph.  That is where a height of 0 occurs