# I need help with a problem dealing with two dimensional projectile motion.

1. Apr 17, 2010

### BrianPagliaci

1. The problem statement, all variables and given/known data
In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower height is h = 10.80 m, and that the bulls-eye's horizontal distance is d = 3.4 m from the launch point. (Neglect air resistance.)

If a jack-o-lantern is given an initial horizontal speed of 3.3 m/s, what are the direction and magnitude of its velocity at the following moments in time? (Let the +x axis point to the right.)

(a) 0.75 s after launch

_____________.° (below the +x-axis)

____________ m/s

(b) just before it lands
_______________ ° (below the +x axis)
______________ m/s

2. Relevant equations
I am not entirely sure but I believe the relevant equations are....
Vot + .5gt^2
d = 3.4m
h= 10.80m
It would be great if someone could give me the required equations and method to solving this kind of problem. Thanks.

3. The attempt at a solution

I am confused how to proceed or all that is required in the solution of this problem.
What I have done so far is substituted into the Vot + .5gt^2 to find vertical and horizontal (x and y) distances. I did this by using the equation given .75 seconds accounting for the initial velocity on the horizontal (3.3 m/s). I am unsure how to proceed though or if i am even doing this right. Very confused.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 17, 2010

### Andrew Mason

The equations you need are:

(1) $$x = v_{0x}t$$ and

(2) $$y = v_{0y} - \frac{1}{2}gt^2$$

In terms of t, (1) can be written:

(3) $$t = x/v_{0x}$$

If the initial motion is only horizontal, v0y=0, so (2) can be written in terms of time, t:

(4) $$t = \sqrt{-2y/g}$$

AM

3. Apr 17, 2010

### GRB 080319B

First, start by drawing a picture of the scenario. The projectile is at a height h above the origin, has initial velocity vector $$v_{0x}$$ in the x-direction, and weight vector mg in the -y-direction. Also, the target is distance d along the x-axis. Start by breaking any forces on the projectile into x and y components. The only force acting on the projectile is the gravitational force.

x component
$$\sum$$$$F_{x}$$ = 0 = m$$a_{x}$$

$$a_{x}$$ = 0 (1)

$$v_{x}$$ = $$\int$$$$a_{x}$$dt = $$v_{0x}$$ (2)

x = $$\int$$$$v_{x}$$dt = $$v_{0x}$$t + $$x_{0}$$ (3) (where $$x_{0}$$ = 0)

y component
$$\sum$$$$F_{y}$$ = -mg = m$$a_{y}$$

$$a_{y}$$ = -g (4)

$$v_{y}$$ = $$\int$$$$a_{y}$$dt = -gt + $$v_{0y}$$ (5)

y = $$\int$$$$v_{y}$$dt = -(1/2)$$gt^{2}$$ + $$v_{0y}$$t + $$y_{0}$$ (6) (where y$$_{0}$$ = h)

We need to find the initial vertical speed $$v_{0y}$$ required to hit the target. We know that when the projectile hits the target, x = 3.4 m and y = 0 m. We can start with equation (3) to find the time of impact when x = 3.4 (t = $$\frac{d}{v_{0x}}$$). At t = $$\frac{d}{v_{0x}}$$, y = 0, so we can solve equation (6) for the initial vertical speed $$v_{0y}$$. We can now use equations (2) and (5) to find the x and y components of velocities at any time. The angle of the velocity vector will be negative wrt the +x-axis (since the projectile is falling) and is given by $$\theta$$ = $$tan^{-1}($$$$\frac{v_{y}}{v_{x}}$$). You can see this by drawing the x and y components of velocity on the diagram, $$tan \theta = \frac{v_{y}}{v_{x}}$$. The magnitude of the velocity vector will be the vector sum of the x and y components v = $$\sqrt{v_{y}^2 + v_{x}^2}$$

Last edited: Apr 17, 2010