A rifle fires a bullet at a speed of 300 m/s at an angle of 42 degrees. What height does the bullet reach? What is the bullet's horizontal range?

Question

I could use some help on exactly how to solve both problem, like what equations/formulas to use and all. For some reason looking at the instructor's notes and how-to explanation actually made me a tad bit more confused. Thanks!

 

 

Answer 1

We have to look at the 300m/s at 42 degrees as a combination of two velocities, a vertical velocity and a horizontal velocity.

 

So, using trigonometry, the vertical velocity is 300m/s * sin 42 degrees and the horizontal velocity is 300m/s * cos 42 degrees.

 

So vertical velocity is 200.7392 m/s  and horizontal velocity is 222.9434 m/s.

 

Now the bullet is going to accelerate due to gravity at -9.8m/s/s

 

So we can find the time it will take for the bullet to lose it's vertical velocity

 

Time = (final velocity - initial velocity)/acceleration = (0 - 200.7392)/-9.8 = 20.48359 seconds.

 

We can assume that the time the bullet will travel on the way down will equal that traveled on the way up,

 

So total time of travel will be 40.96718 seconds.

 

We can find the maximum height by using distance = initial velocity * time = 1/2 * acceleration * square of time

 

So the maximum height is 200.7392*20.48359 + .5*-9.8*20.48359^2 = 2055.93 meters

 

And we can do the same with the horizontal range, using total travel time and 0 acceleration

 

So the horizontal range is 222.8434*40.96718 + .5*0*40.96718 = 9133.364 meters.

 

Given the degree of accuracy in our data we should say that the bullet will travel to a height of approximately 2056 meters and a range of approximately 9133 meters.