physics homework help

Question

A circus performer is shot out of a cannon and flies over a vertical net that is places at a horizontal distance of 8.80m from the cannon. When the cannon is aimed at an angle of 40 degrees above the horizontal, the performer is moving in the horizontal direction and just barely clears the net as he passes over it. a) what is the muzzle speed of the cannon m/s? and b) find the height of the net m?

Answer 1

First: Draw a picture of the situation. Everything here can be represented and calculated in 2 dimensions. If we imagine viewing the act from the right side of the cannon, then we will see the cannon facing to the right from our perspective, with the net acting like a wall of unknown height 8.80 m to the right of the cannon. We will use "right" as the positive-x direction and "up" as the positive-y direction. The cannon can be the point (0,0).

We've been tasked with finding the muzzle velocity of the cannon (that's just the initial velocity of the performer) and the height of the net. The whole "the performer is moving in the horizontal direction and just barely clears the net" thing is just another way of saying the following:

-The performer's trajectory has a maximum height which is exactly the height of the net

-The performer's trajectory has a maximum height which is horizontally aligned with the net

In other words, the coordinates of the performer's position when clearing the net are (x_n, y_n), and y_n is the height of the net, while x_n = 8.80, the horizontal distance of the net from the cannon.

We need to use a formula that has what we need to find (the max height of the performer and their initial velocity) as well as what we already have. Here's one that seems like it will help:

a = (v_f - v_i) / t_f

Remember that this can only be used when acceleration is constant.

In the x-direction, this doesn't help us. There is no air resistance, so a in the x-direction is 0. This just means v_i = v_f, but we don't know v_f.

In the y-direction, it may be a little more useful. We know a is about -9.81 m/s², because the only acceleration is from gravity (and we made "up" the positive-y direction, so gravity is negative). This shows that acceleration is constant, so it's OK to use constant-a formulae. We also know v_f if we're talking about the moment the performer clears the net-- they are "moving in the horizontal direction" at this point, meaning the y-component of their velocity is 0. So now we have:

-9.81 = (0 - v_iy) / t_f = -v_iy / t_f

We don't know t_f yet, so we are left with 2 unknowns and we can't get any further. Let's keep searching... How about this one?

r(t) = 1/2at² + v_it + r_i = 1/2at² + v_it

Note that r_i is (0,0), so we can just drop it.

In the x-direction, there is no acceleration, so this simplifies to the familiar

x(t) = v_ixt

Since we're mainly interested in what happens at x(t_f), which is just 8.80 m:

x(t_f) = v_ixt_f = 8.80 m

Two unknowns again. But clearly, t_f is an important quantitiy. So what about the y-direction? Again, the y acceleration is just -9.81 m/s², so

y(t) = (1/2)(-9.81)t² + v_iyt = -4.905t² + v_iyt

The value of y at t_f is y_n (height of the net!), so:

y(t_f) = y_n = -4.905t_f² + v_iyt_f

Blast it. This one is three unknowns. However, we have at least set up a lot of relationships that should come in handy if we can start to crack this case.

We remember that if there are two unknowns, we can solve for both of them if we have two different equations in those same two unknowns. But sadly, looking back, all of our different equations have different unknowns. The closest we get are these two:

v_ixt_f = 8.80 m

-v_iy / t_f = -9.81 m/s²

because they both have t_f and a component of v_i, but they each use different components, one x and one y. So it's a no-go so far.

...But v_ix and v_iy are the two components of the same vector, right? Aren't they already related somehow? Yes they are! As with all 2D vectors, if we draw the vector v_i as the hypotenuse of a right triangle with the x- and y-components as the legs, noting that the cannon is tilted at 40° above the horizontal, we can use simple trig ratios to relate all the parts. When drawn correctly, the x-component is the leg adjacent to the known angle, and the y-component is opposite. Consult the trig chant to discover which ratio uses the sides we know:

sOH cAH tOA

The winner is tOA, which just means

tan(θ) = O/A

or

A tan(θ) = O

If we change the names of the parts to those in our problem and sub in our angle it's this:

v_ix tan(40°) = v_iy

Now we're getting somewhere. Let's bring down our equations from before:

v_ixt_f = 8.80 m

-v_iy / t_f = -9.81 m/s²

Substitute our new expression for v_iy:

v_ixt_f = 8.80 m

-(v_ix tan(40°)) / t_f = -9.81 m/s²

Now we finally have two equations of two unknowns. We can see where this is going... let's solve for one of these variables:

-(v_ix tan(40°)) / t_f = -9.81 m/s²

v_ix = (9.81 / tan(40°)) t_f m/s

Substitute this expression for v_ix into the other equation:

v_ixt_f = 8.80 m

((9.81 / tan(40°)) t_f m/s)t_f = 8.80 m

(9.81 / tan(40°))t_f² = 8.80 s²

t_f = √(0.9tan(40°)) s = 0.87 s

NOTE: We discount the negative square root, since negative time is meaningless in this problem.

We have found the all-important t_f. We can take our pick from many of the above equations in order to reach more solutions. I choose this one:

-v_iy / t_f = -9.81 m/s²

-v_iy / (0.87 s) = -9.81 m/s²

v_iy = 8.53 m/s

Now that we have one component of the initial velocity, we can find its magnitude. There are a few ways we can get there, but I will just use another trig ratio. Recall from before that v_i is the hypotenuse, and v_iy is the leg opposite the known angle. The ratio we want is

sOH

or

sin(θ) = O/H

sin(40°) = v_iy / v_i

v_i = v_iy / sin(40°) = (8.53) / sin(40°) m/s = 13.27 m/s

This is the answer to (a).

Now for the net height. From waaay above:

y_n = -4.905t_f² + v_iyt_f = -4.905(0.87)² + (8.53)(0.87) m = 3.71 m

This is the answer to (b).