At what distance would a net need to be placed to catch the performer at an equal height to the height she left the cannon?

Question

A circus stunt performer is shot from a cannon tilted at 42.0 ° with an initial velocity of 22.5 m/s.At what distance would a net need to be placed to catch the performer at an equal height to the height she left the cannon?

Answer 1

This problem is a two step process for high school students, who are expected to show the steps using first principles. This is a typical projectile motion (one that has a horizontal and vertical component of motion) problem.

(I) Find the time, t, taken for the flight by considering the vertical part of the projectile motion.

(II) Find the horizontal distance travelled (also called the range, in this case) by using the horizontal component of the motion.

(I) In this step, consider only the vertical motion. Since the net will have to be placed at the same height as the cannon, we know that the height difference between the initial and final position (Delta y) =0.

Use the second kinematic equation to find the time, t.

(Delta y) = v_yt + (1/2)at²

Substituting for (Delta y) and rearranging the equation, we get;

t = (-2v_y)÷a

Since we are considering vertical motion, a = -g, the acceleration due to gravity. We are taking the downward direction as negative, which is why acceleration is negative.

Also, v_y = vsinθ, the vertical or the y-component of the initial velocity. [See the chapter on vectors and projectile motion if you forgot how to resolve vectors into components].

t = (-2vsinθ)÷(-g)

Or, t = (2vsinθ)÷(g) [the negatives cancel off]

Or, t = (2*22.5(m/s)*sin(42.0^o)÷(9.8m/s²) = 3.07 seconds.

(II) Use the same kinematic equation as above (with subscripts changed to x) to find the range. Note that here, the unknown is (Delta x) and the horizontal acceleration is zero (in every projectile motion, the only force acting on an object is gravity, which is in the vertical direction; there are no forces and accelerations in the horizontal direction). The time taken remains the same in both horizontal and vertical motions.

Also, v_x = vcosθ, the horizontal component of initial velocity.

(Delta x) = v_xt + (1/2)at²

(Delta x) = (vcosθ)t + (1/2)a_xt²

(Delta x) = (vcosθ)t + 0 [ because, a_x = 0]

(Delta x) = (22.5(m/s)*cos(42.0^o))*3.07(s)

The answer for range will be in meters.

The answer is lengthy, but this is what teachers expect students to answer, especially in AP, and even in regular physics classes.

Answer 2

Range= U² Sin 2 Theta/ g

Here U = 22.5m/s Theta = 42.0 , 2Theta = 84

Plug in

Range = (22.5)^{2 *} (Sin 84)/ 9.8

This answer is in m