Velocity of a Projectile, Help!!

[Guest]

I need help getting on track with this Velocity problem, so here it is!

1. Two watermelons are fired at the same time from the same location in opposite directions. The first projectile is fired at a 30° angle to the ground, and the second is fired at a 45° angle. The velocity of the first projectile is 160 feet per second, and the velocity of the second watermelon is feet per second.

a. Write an equation representing the vertical distance between the two watermelons as a function of time, dy(t), and one for the horizontal distance, dx(t). (You don’t need the distance formula for this part.)
b. Write an equation representing the total distance between the two watermelons as a function of time, d(t). (Use the distance formula here.)
c. Find derivatives of dx(t), dy(t), and d(t). Explain what they mean in the context of the problem, and describe how the distance between the watermelons changes with time. You may want to include graphs.
d. Find d’(5) and explain what it means in the context of the problem.

I also have these helpful hints/formulas ...
The height of a projectile fired straight up from ground level with a velocity of v0 ft/sec is given by
y = v0t – 16t2
If the projectile is fired at an angle to the ground instead of straight up, use the vertical component of v0
y = v0*sin(?)*t – 16t2
A projectile fired at an angle to the ground will also move in the x-direction. The distance is given by
x = v0*cos(?)*t
Can someone please help explain to me where I am going with this information? I wanted to solve the problem trigonometrically at first but I didn't get what I was looking for

 

[tkhunny]

That does look like you have enough information.

What did you do and how close was it? Please demonstrate.

 

[soroban]

Hello, KgMarie13!

You left out a piece of information . . . but I'll get you started.

1. Two watermelons are fired at the same time from the same location in opposite directions.
The first projectile is fired at a 30° angle to the ground, and the second is fired at a 45° angle.
The velocity of the first projectile is 160 feet per second,
. . and the velocity of the second watermelon is ??? feet per second. . Call it v.

a. Write an equation representing the vertical distance between the two watermelons as a function of time, dy(t)
. . and one for the horizontal distance, dx(t). (You don’t need the distance formula for this part.)

\(\displaystyle \text{For the 1st melon, the equations are: }\;\begin{Bmatrix} x_1 &=& (160\cos30^o)t &=& 80\sqrt{3}\,t \\ y_1 &=& (160\sin30^o)t - 16t^2 &=& 80 - 16t^2 \end{Bmatrix}\)

\(\displaystyle \text{For the 2nd melon, the equations are: }\;\begin{Bmatrix}x_2 &=& (v\cos45^o)t &=& \frac{v}{\sqrt{2}}t \\ y_2 &=& (v\sin45^o)t - 16t^2 &=& \frac{v}{\sqrt{2}}t - 16t^2 \end{Bmatrix}\)


\(\displaystyle \text{Their vertical distance is:}\)
. . \(\displaystyle dy(t) \:=\:y_2 - y_1 \:=\:\left(\frac{v}{\sqrt{2}}\,t - 16t^2\right) - \left(80t - 16t^2\right) \;=\;\left(\frac{v}{\sqrt{2}} - 80\right)t\)

\(\displaystyle \text{Their horizontal distance is:}\)
. . \(\displaystyle dx(t) \:=\:x_1-x_2 \:=\:80\sqrt{3}\,t - \frac{v}{\sqrt{2}}\,t \;=\;\left(80\sqrt{3} - \frac{v}{\sqrt{2}}\right)t\)

 

[Guest]

OK, the parts that I forgot...
The distance formula,
gives the distance between two points in an xy-plane, (x1, y1) and (x2, y2).
The velocity of the second melon, 128root2 feet per second.

Thank you for getting me this far! I might still need more guidance :?
Most importantly, how do I put equations into my posts?

 

[JeffM]

Most importantly, how do I put equations into my posts?

If you mean how do you put mathematical notation into your posts, you can make it look professional by learning LaTex, which is what Soroban uses. It is not particularly easy, but you can take a look at what it looks like by pushing the quote button on Soroban's reply to you in this thread.

On the other hand, you can be lazy like me. Post fractions as x / y. Show subscripts like a[sub:jvx5gvk8]1[/sub:jvx5gvk8]. You do that by typing a1, highlighting the 1, and clicking the "sub" button on the toolbar above the writing space. You can show superscripts for exponents, example 2[sup:jvx5gvk8]lnx[/sup:jvx5gvk8], by highlighting lnx, and clicking the sup button. You can also show exponents using the caret (shift key and 6) as in x^2 for x[sup:jvx5gvk8]2[/sup:jvx5gvk8]. What is very important when you do things the lazy way (that is my way) is to use parentheses carefully. If you write y = 2 / x + 1, no one will know if you mean y = [(2 / x) + 1] or y = [2 / (x + 1)].

You can show "greater or equal" as >=. You can show "leads to" or approaches as -->. Where my laziness becomes a problem is with integrals, limits, and derivatives. If you do not know Latex, you have to use a lot of words: like the integral from a to b of f(x)dx. That is hard for people to comprehend because math notation is what people are familiar with.

 

[Guest]

Thank you all for the helpful help : ) BUT I have 1 more question though!
Based on my first post, I'm working through a,b,c, and I am stuck on 'd'. Well I guess I may have done 'c' wrong... (I will try to make this as clear as possible)
For part 'c' I started with the three equations and found their derivatives; Dx(t)=10.564t ---so the derivative is--- D'x(t)=10.564
Dy(t)=48t ---so the derivative is--- D'y(t)=48
D(t)=49.149t ---so the derivative is--- D'(t)=49.149
I am not sure but I don't think my derivatives should be horizontal and vertical lines... right? Or are the derivatives giving me the maximum height of each melon? ie. the max height of the 1st melon is 10.564 ft? I might be crazy...
Did I make a mistake in part 'a' and 'b' with the vertical and horizontal equations? * I used the information given by Soroban and the missing velocity.