I with a problem dealing with two dimensional projectile motion.

[BrianPagliaci]

Homework Statement

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

Homework 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.

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.

 

[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

 

[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
\sumF_{x} = 0 = ma_{x}

a_{x} = 0 (1)

v_{x} = \inta_{x}dt = v_{0x} (2)

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

y component
\sumF_{y} = -mg = ma_{y}

a_{y} = -g (4)

v_{y} = \inta_{y}dt = -gt + v_{0y} (5)

y = \intv_{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}