# Question!!!!!

• Jun 1st 2006, 01:40 PM
kkmarie9009
Question!!!!!
heres the question

A rock is thrown vertically from the ground with a velocity of 24 meters per second and it reaches a height of 2+24t-4.9t(Squared) after t seconds. How many seconds after the rock is thrown will it reach maximum height and what is the maximum height the rock will reach in meters? how many seonds after the rock is thrown will it hit the ground? Round your answers to the nearest hundredth

• Jun 1st 2006, 03:05 PM
Jameson
First thing is I don't agree that this is thrown from the ground. At t=0, the position is 2. So I guess we can call 2 meters the origin. The rock will reach its max height at the vertex of this parabola. Have you taken Calculus?
• Jun 1st 2006, 03:11 PM
kkmarie9009
no I haven't taken that......Im in Math B in NYS
• Jun 1st 2006, 04:10 PM
Soroban
Hello, kkmarie9009!

Quote:

A rock is thrown vertically from the ground with a velocity of 24 m/sec
and it reaches a height of $h \,= \,2 + 24t - 4.9t^2$ after t seconds.

(a) How many seconds after the rock is thrown will it reach maximum height?
(b)What is the maximum height the rock will reach in meters?
(c) How many seconds after the rock is thrown will it hit the ground?
The function is a quadratic (second degree).
. . Its graph is a down-opening parabola.
. . The maximum height occurs at the vertex of the parabola.
The formula for the vertex of a parabola is: $x = \frac{-b}{2a}$

Our parabola has: $a = -4.9,\;b = 24,\;c = 2$
The vertex is at: $t \:= \:\frac{-24}{2(-4.9)} \:=\:\frac{120}{49} \:= \:2.4489795$

(a) Therefore, the rock reaches maximum height at $t = 2.45$ seconds.

If $t = \frac{120}{49}$, then: $h\;=\;2 + 24\left(\frac{120}{49}\right) - 4.9\left(\frac{120}{49}\right)^2 \;=\;31.3877551$

(b) Therefore, the maximum height is: $h \:= \:31.39$ m.

If the rock hits the ground, its height is zero.
. . Solve the quadratic: $2 + 24t - 4.9t^2\:=\:0\quad\Rightarrow\quad 4.9t^2 - 24t - 2\:=\:0$

Quadratic Formula: $t \;= \;\frac{-(-24) \pm \sqrt{(-24)^2 - 4(4.9)(-2)}}{2(4.9)}$

Hence: $t\;= \;\frac{24 \pm \sqrt{615.2}}{9.8} \;= \;4.979920979$

(c) Therefore, the rock hits the ground in about $4.98$ seconds.
• Jun 1st 2006, 04:26 PM
kkmarie9009
Quote:

Originally Posted by Soroban
Hello, kkmarie9009!

The function is a quadratic (second degree).
. . Its graph is a down-opening parabola.
. . The maximum height occurs at the vertex of the parabola.
The formula for the vertex of a parabola is: $x = \frac{-b}{2a}$

Our parabola has: $a = -4.9,\;b = 24,\;c = 2$
The vertex is at: $t \:= \:\frac{-24}{2(-4.9)} \:=\:\frac{120}{49} \:= \:2.4489795$

(a) Therefore, the rock reaches maximum height at $t = 2.45$ seconds.

If $t = \frac{120}{49}$, then: $h\;=\;2 + 24\left(\frac{120}{49}\right) - 4.9\left(\frac{120}{49}\right)^2 \;=\;31.3877551$

(b) Therefore, the maximum height is: $h \:= \:31.39$ m.

If the rock hits the ground, its height is zero.
. . Solve the quadratic: $2 + 24t - 4.9t^2\:=\:0\quad\Rightarrow\quad 4.9t^2 - 24t - 2\:=\:0$

Quadratic Formula: $t \;= \;\frac{-(-24) \pm \sqrt{(-24)^2 - 4(4.9)(-2)}}{2(4.9)}$

Hence: $t\;= \;\frac{24 \pm \sqrt{615.2}}{9.8} \;= \;4.979920979$

(c) Therefore, the rock hits the ground in about $4.98$ seconds.

Exactly why would you use the quadratic Formula?? is there an easier way to figure it out without it? or no?
• Jun 1st 2006, 05:52 PM
Quick
Quote:

Exactly why would you use the quadratic Formula?? is there an easier way to figure it out without it? or no?
The Quadratic formula may seem long but it is pretty simple to do on a calculator. However, there are two other possible ways to solve quadratic functions. (I'm using example equations)

Factoring:
$2x^2+7x+3=y$ Start with equation
$(2x+1)(x+3)=y$ Factor
$(2x+1)(x+3)=0$ Substitute 0 for "y"
$2x+1=0$ or $x+3=0$ anything multiplied by 0 equals 0, so solve each factor for 0.
$x=-0.5$ or $x=-3$ Solve

Completing the Square:
$x^2+7x+3=y$ Start with equation
$x^2+7x=-3$ subtract "c" from both sides
$x^2+7x+12.25=9.25$ add $\left(\frac{b}{2}\right)^2$ to both sides
$(x+3.5)^2=9.25$ Factor
$\sqrt{ (x+3.5) }^2= \pm \sqrt{ 9.25 }$ Find the square root of both sides
$x+3.5= \pm 3.014381$ Solve square roots
$x+3.5=3.01$ or $x+3.5=-3.01$ Solve each answer
$x=-.48$ or $x=-6.51$ Solve