Equal-Range firing angles - What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun if the projectiles initial speed is 400 m/sec?

This one needs to be done in two parts. If you have taken any physics courses, you should know the formula for position is \[x = vt+1/2at^2\]. Given that the units of measurement are in metric, we can knock this out pretty easy. First we need to find out how fast the projectile will travel the 16km with an initial speed of 400m/s. We solve for x here. \[16km = 400(m/s)t-4.9(m/s^2)t^2\], where t= time in seconds. The -4.9 is 1/2 times the gravitaional acceloration of 9.8 meters per second squared. Now that we have the equation, we can solve for t using the quadratic formula. \[400(m/s)\pm \sqrt{400^2-4(-4.9m/s^2)(16m)}\div2(4.9)\] Doing the math here, we find that t= 81.5926, and .04 seconds. Now we can solve for the y coordinate, which is when the projectile will hit at the same elevation as the gun. Since we need an elevation, we need an angle, which means that we will be using the sine function. \[y= 400\sin \theta t-4.9t^2\] Simply plug in the numbers we got for time and then solve for sine by using the inverse sine function. you should get 88.2026 and .028075 degrees

Generally, for calculus based physics courses or math classes, you may deal in radians instead, which you can either set your calculator to rad, or use the degrees to radians formula, which is \[degrees \times(180/\pi)\]

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