# At a basketball game, members of the spirit club use a t-shirt cannon to launch team t-shirts to spectators in the stands.

At a basketball game, members of the spirit club use a t-shirt cannon to launch team t-shirts to spectators in the stands. the equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v. h = -16t 2 + vt + c solve the equation for initial velocity, v. the cannon launched a t-shirt from an initial height of 5 feet. after 3 seconds the shirt has reached a height of 131 feet. use your equation from 1 to find the initial velocity of the t-shirt. show your work. what does the equation from 1 tell you about the value of the velocity, v, when t = 0? explain.

## This Post Has 10 Comments

1. pbenavid9849 says:

The equation for initial velocity is s = ut + 1/2at^2, where s = distance, u = initial velocity, and a = acceleration.

Step-by-step explanation:

2. bairdmatthew43 says:

The two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

Step-by-step explanation:

As the given equation is

$y''+7y'-8y=0\\$

So the corresponding equation is given as

$m^2+7m-8=0$

Solving this equation yields the value of m as

$(m+8)(m-1)=0\\m=-8, m=1$

Now the equation is given as

$y(t)=C_1e^{m_1t}+C_2e^{m_2t}$

Here m1=-8, m2=1 so

$y(t)=C_1e^{-8t}+C_2e^{t}$

The derivative is given as

$y'(t)=-8C_1e^{-8t}+C_2e^{t}$

Now for the first case y(t=0)=1, y'(t=0)=0

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=1\\-8C_1+C_2=0$

Solving the equation yield

$C_1=1/9 \\C_2=8/9$

So the function is given as

$y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$

Now for the second case y(t=0)=0, y'(t=0)=1

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=0\\-8C_1+C_2=1$

Solving the equation yield

$C_1=-1/9 \\C_2=1/9$

So the function is given as

$y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

So the two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

3. valdezangie10 says:

1. h = -16t² + vt + c
⇒ vt = 16t² + h - c
⇒ v = 16t + (h-c)/t

2. c = 5 ft
t = 3s
h = 131 ft

v = 16*3 + (131-5)/3 = 48 + 42 = 90 ft/s

when t=0, velocity is not defined. That means at t=0, there is no velocity of the t-shirt or it is 0.

4. nandalabella06 says:

a. $P = 1000 - Ce^{-\frac{kt^2}{2} }$

b. $C = 950$

c. $P = 1000 - 950e^{-\frac{kt^2}{2} }$

Step-by-step explanation:

a. Let the number of penguins who have the disease t days after the outbreak be P

Initial number of penguins = 1000

Therefore, current number of penguins = 1000 - P

And the rate of spread of disease according to the statement is

$\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)$

where k is the constant of proportionality

$\frac{dP}{1000-P}=kt.dt$

Integrating both sides

$-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }$

b. Seeing as 50 penguins had the disease initially,

t = 0

P = 50

The general solution of the differential solution becomes

50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)

$C = 1000 - 50 = 950$

c. Therefore, the solution that satisfies the initial condition is

$P = 1000 - 950e^{-\frac{kt^2}{2} }$

5. nerual0503 says:

A) Judging by the slope field, it would appear the solution passing through (0, 2) would just be the line $y=2$. This holds up for the given ODE, since if $y(x)=2$, then both sides of the ODE reduce to 0.

Since we can surmise that $y=2$ is an equilibrium for this ODE (that is, the derivative along this line is 0 everywhere), and the slope at (1, 0) is positive, we know that as $x\to+\infty$, the function $y(x)$ will converge to 2. In other words, as $x$ gets larger, the slope field suggests that the solution curve through (1, 0) will start to plateau and steadily approach $y=2$. On the other hand, as $x\to-\infty$, the slope field tells us that the curve would rapidly diverge off to $-\infty$. (When you actually draw the solution, you would end up with something resembling the plot of $-e^{-x}$.)

b) The tangent line to $y=f(x)$ at $x=1$, given that $f(1)=0$, takes the form

$\ell(x)=f(1)+f'(1)(x-1)$
$\ell(x)=f'(1)(x-1)$

When $x=1$, we have $y=0$, so

$f'(1)=\dfrac13\cdot1\cdot(0-2)^2=\dfrac43$

and so the tangent line to $f(x)$ at $x=1$ is

$\ell(x)=\dfrac43(x-1)$

Using the tangent line as an approximation, we would find

$f(0.7)\approx\ell(0.7)=\dfrac43\left(\dfrac7{10}-1\right)=-\dfrac4{10}=-0.4$

c) The ODE is separable, so we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac13x(y-2)^2\iff\dfrac{\mathrm dy}{(y-2)^2}=\dfrac x3\,\mathrm dx$

Integrating both sides gives us

$-\dfrac1{y-2}=\dfrac{x^2}6+C$

Given that $y(1)=0$, we get

$-\dfrac1{0-2}=\dfrac{1^2}6+C\implies C=\dfrac13$

so the particular solution is

$-\dfrac1{y-2}=\dfrac{x^2}6+\dfrac13$

You're asked to find the solution in the form $y=f(x)$, so you should solve for $y$. You would end up with

$y=-\dfrac6{x^2+2}+2=\dfrac{2(x^2-1)}{x^2+2}$

6. Thania3902 says:

h(t) = -16t^2 + vt + 5   h(3) = 131 ft   h(3) = -16(3)^2 + 3v + 5 = 131   3v = 270   v = 90 ft/s

7. french31 says:

1.

$h=-16t^2+vt+c\qquad\text{original equation}\\\\h+16t^2-c=vt\qquad\text{subtract terms not containing v}\\\\v=\dfrac{h+16t^2-c}{t}\qquad\text{divide by the coefficient of v}$

2. Filling in the given numbers, we have ...

$v=\dfrac{131+16(3)^2-5}{3}=90\qquad\text{ft/s}$

3. The velocity is undefined when t=0. The equation is telling you it would take infinite velocity to travel some distance in zero time. The equation is meaningless in such a situation.

8. cecilysimpson3089 says:

Initial velocity is 3.5. The equation is s = ut + 1/2at^2, where s - distance, u - inititial velocity, and a - acceleration. How do I modify the acceleration formula to one that gives me initial velocity? You can't change the acceleration formula to one that gives you the initial velocity you want, as a=v/t.

Step-by-step explanation:

9. glander2 says:

h(t) = -16t^2 + vt + 5   h(3) = 131 ft   h(3) = -16(3)^2 + 3v + 5 = 131   3v = 270   v = 90 ft/s

10. AnkitDavid1616 says:

The required formula is $v=\frac{h-c}{t}+16t$. The initial velocity is 90 ft/s. At t=0, the value of v is undefined.

Step-by-step explanation:

The equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v.

$h=-16t^2+vt+c$

We need to solve this equation for v.

Subtract c from both sides.

$h-c=-16t^2+vt$

$h-c+16t^2=vt$

Divide both sides by t.

$\frac{h-c+16t^2}{t}=v$

$v=\frac{h-c}{t}+16t$         .... (1)

Therefore the required formula is $v=\frac{h-c}{t}+16t$.

We need to find the velocity for

Initial height c = 5 feet

Time t = 3 sec

Height h = 131 feet

Substitute c = 5, t = 3 and h = 131 in equation (1).

$v=\frac{131-5}{3}+16(3)$

$v=42+48=90$

Therefore the initial velocity is 90 ft/s.

Substitute t=0 in equation (1).

$v=\frac{h-c}{0}+16(0)$

$v=\infty$

At t=0, the value of v is undefined.