Question

A gun has a muzzle speed of 80 meters per second. What angle of elevation should be used to hit an object 180 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.

The answer must be in radians.

I found: angle=1/2arcsin (180*9.8)/(80^2), but it was not correct.

Answer 1

The way to tackle this problem is to understand it as components of Vertical and Horizontal velocities. First,

The vertical distance can be defined in terms of it's verticle velocity & the acceleration applied, as shown below:

\[s = s_o + v_o \times t + (a \times t^2 / 2)\]

Initial vertical distance, \[s_o = 0m.\] Final vertical distance, \[s = 0m.\]

The assumption is that the object has the same elevation as the the spot where the gun is being fired. Initial vertical velocity is \[v_v = 80 \times \sin \theta\], so the above equation becomes \[0 = 80 \times \sin \theta \times t + 4.9 \times t^2 ...... (1)\]
The distance horizontally to be covered is 180m. Since the assumption of no drag being applied then there is no deceleration in this case, this gives \[v_h = 80 \cos \theta\] Substitute this into the top equation we get: \[180 = 80 \cos \theta \times t\] Rearrange and we get: \[t = 180 / (80 \cos \theta) ............ (2)\] Substitute (2) into (1) you get \[(24.8 / \cos^2 \theta) + (180 \sin \theta / \cos \theta) = 0\] Multiply the right term by \[\cos \theta / \cos \theta\] and rearrange you get \[\sin \theta \cos \theta = -24.8/180\ ........(3)\] Now \[\sin (2\theta) = 2 \sin \theta \cos \theta\] so equation (3) becomes \[\sin(2 \theta) = -49.6/180\] Solve and you get\[\theta = -0.1396 radians\] i.e. the gun is firing to the left at 8 degrees from the horizon.

Answer 2

I need to correct what I wrote. Some of the signs are incorrect in the working out. It should be 0.1396radians if you are going from left to right. I'm not sure what I did wrong. I'm sure you can work it out.