Help with two 2D motion problems, PLEASE

[myelevatorbeat]

I'm almost done with my homework for the week but I'm stuck on these two questions. I've stated all the work I've done thus far below the problems. Any help would be greatly appreciated!


Problem #1:

A projectile is fired with an initial speed Vo at an angle theta to the horizontal. When it reaches its peak, the projectile has (x,y) coordinates given by (R/2, h) and when it strikes the ground its coordinates are (R, 0), where R is called the horizontal range.

a)Show the projectile reaches a maximum height given by:

h= Vo^2sin^2theta / 2g

b) Show the horizontal range of the projectile is given by:

R=Vo^2sin2theta / g





Problem #2:


A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance "X" away from the gun. The pellet hits the target a vertical distance "Y" below the gun.

a) Show that the position of the pellet when traveling through air is given by y=Ax^2, where A is constant.
b)Express the constant A in terms of the initial (muzzle) velocity and the free-fall acceleration.
c)If x=3.00 m and y=0.210 m, what is the initial speed of the pellet?

I already finished part A and got y(t)= -1/2g[x/(Vo)x)]^2.

Now, I need help starting part B, because I'm completely clueless. I figure I'll just put x and y in part C in the equation I get for part B and that should solve for the last part?

 

[Mindscrape]

Which problem did you get, and which one do you need help with? Can you please put up some work, so we can see how far along you are?

 

[m2003]

I would be happy to help you with these two 2D motion problems. Let's start with problem #1. To solve this problem, we can use the equations of motion for projectile motion. These equations are:

1) x = V0 * cos(theta) * t
2) y = V0 * sin(theta) * t - (1/2) * g * t^2
3) Vx = V0 * cos(theta)
4) Vy = V0 * sin(theta) - g * t

Where:
- x and y are the coordinates of the projectile at any given time t
- V0 is the initial velocity
- theta is the angle at which the projectile is fired
- g is the acceleration due to gravity (9.8 m/s^2)

Now, let's use these equations to solve part a of the problem. We know that at the peak of the projectile's motion, its vertical velocity (Vy) will be equal to 0. Therefore, we can set equation 4 equal to 0 and solve for t:

0 = V0 * sin(theta) - g * t
t = V0 * sin(theta) / g

Next, we can substitute this value of t into equation 2 to find the maximum height (h) of the projectile:

h = V0 * sin(theta) * (V0 * sin(theta) / g) - (1/2) * g * (V0 * sin(theta) / g)^2
h = (V0^2 * sin^2(theta)) / 2g

This matches the equation given in the problem, so we have successfully solved part a.

For part b, we can use the same approach. At the horizontal range, the projectile's vertical position (y) will be equal to 0. So we can set equation 2 equal to 0 and solve for t:

0 = V0 * sin(theta) * t - (1/2) * g * t^2
t = (2 * V0 * sin(theta)) / g

Now, we can substitute this value of t into equation 1 to find the horizontal range (R) of the projectile:

R = V0 * cos(theta) * ((2 * V0 * sin(theta)) / g)
R = (2 * V0^2 * sin(theta) * cos(theta)) / g
R