What is the exit velocity of the blowgun dart? If the blowgun is 1.25 meters long, what is the acceleration of the dart inside the blowgun?

Question

A blowgun fired horizontally will strike a target 55 meters away after dropping 9 centimeters. What is the exit velocity of the blowgun dart? If the blowgun is 1.25 meters long, what is the acceleration of the dart inside the blowgun?

Answer 1

Start by figuring out what your knowns are and translating them into the standard variables for kinematics. The gun is fired horizontally, so that means all of the initial velocity is horizontal (x) and so the vertical (y) component must be zero. Remember that for 2d kinematics, you can apply the 1d kinematic equations in each direction independently. In this problem we are talking about the period of time starting just after the dart leave the gun, and just before it hits an object 55 m away, since that is when we have constant acceleration (in this case, acceleration due to gravity).

Let's look at the y-direction motion first. The dart has dropped by 9 cm. We can use y = y0 + v0t + (1/2)at^2 to solve for how long the dart was flying in the air. Use algebra and plug in y = 9 cm = 0.09 m, y0 = 0 cm (we can choose the initial height to be zero), v0 = 0 m/s, a = 9.8 m/s^2. Now you can solve for t to get t = sqrt(2y/a) = 0.1355 s or about 0.14 s.

Once we have t, we can solve for the initial velocity exiting the gun, which would be 55 m/0.1355 s = 406 m/s. Recall that the horizontal velocity is constant.

If the blowgun is 1.25 m long, we can solve for the acceleration using v^2 = v0^2 + 2a(x-x0). The dart presumably starts from rest, so v0 = 0 m/s. Also v = 406 m/s from above. x-x0 = 1.25 m. Solve for a = (v^2 - v0^2)/(2*(x-x0)), or about 66,000 m/s^2.