• Feb 28th 2008, 08:49 AM
inordinatemuse
Okay, so here is the question:

When an object such as a bullet or a ball is shot or thrown upward with an initial velocity, its height is given by a quadratic function.

s(t)= -4.9tē+Vot+h

In this function, h is the starting height in meters, Vo is the velocity, s is the actual height (also in meters) and t is the time from projection in seconds. This model is based on the assumption that there is no air resistance and that the force of gravity pulling the object earthward is constant. Neither of these conditions exists precisely, so this model (as is the caSe with most mathematical models) gives only approximate results. A model rocket is fired upward. At the end of the burn, it has an upward velocity of 40m/second and is 155m high. Find (a) its maximum height and when it is attained and (b) when the rocket reached the ground.

Hint: 1) Vertex will give max height.
2) s(t) is 0 when rocket reaches ground

• Feb 28th 2008, 09:10 AM
TheEmptySet
Quote:

Originally Posted by inordinatemuse
Okay, so here is the question:

When an object such as a bullet or a ball is shot or thrown upward with an initial velocity, its height is given by a quadratic function.

s(t)= -4.9tē+Vot+h

In this function, h is the starting height in meters, Vo is the velocity, s is the actual height (also in meters) and t is the time from projection in seconds. This model is based on the assumption that there is no air resistance and that the force of gravity pulling the object earthward is constant. Neither of these conditions exists precisely, so this model (as is the caSe with most mathematical models) gives only approximate results. A model rocket is fired upward. At the end of the burn, it has an upward velocity of 40m/second and is 155m high. Find (a) its maximum height and when it is attained and (b) when the rocket reached the ground.

Hint: 1) Vertex will give max height.
2) s(t) is 0 when rocket reaches ground

$s(t)=-4.9t^2+40t+155$

use the vertex formula to find the max

$t=\frac{-b}{2a}$

after getting the t coordinate plug back into the equation to find the height. Setting the equation equal to zero and solving will answer part two.. ie.

$0=-4.9t^2+40t+155$ I would use the quadratic formula...

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

P.S. does anyone know the LaTex code for the plus or minus symbol..

thanks
• Feb 28th 2008, 09:24 AM
xifentoozlerix
Quote:

Originally Posted by TheEmptySet
P.S. does anyone know the LaTex code for the plus or minus symbol..
thanks

\pm