How Far Did Big Bertha's Shell Travel?

[steelblondie5]

Homework Statement

During World War 1, the Germans had a gun called Big Bertha that was used to shell Paris. The shell had an initial speed of 2.26 km/s at an initial inclination of 65.8 degrees to the horizontal. The acceleration of gravity is 9.8 m/s^2. How far away did the shell hit? Answer in units of km.
How long was it in the air? Answer in units of s.

Homework Equations

Vf= Vi + a (tf-ti)
y= yi + vit + 1/2 at^2.

The Attempt at a Solution

I did it several different ways and keep getting answers that are wrong. I got 4.10 s for time, then I got .420s and .840 s. I keep getting the wrong answer. Once I figure out how to do time, I'll be able to do the distance part. Keep in mind the a is in meters per second but kilometers is required in the answer.

 

[edziura]

Which of the two eqns did you use to find the time of flight t? What values did substitute?

 

[baba89]

I would suggest approaching this problem by breaking it down into smaller steps and using the appropriate equations and units for each step. First, convert the initial speed of the shell from 2.26 km/s to m/s. This can be done by multiplying 2.26 km/s by 1000 to get 2260 m/s.

Next, use the given initial inclination of 65.8 degrees to find the initial vertical and horizontal components of the velocity using trigonometry. The initial vertical component would be 2260 m/s * sin(65.8) = 2034.8 m/s, and the initial horizontal component would be 2260 m/s * cos(65.8) = 1095.9 m/s.

Then, use the equation y = yi + vit + 1/2 at^2 to solve for the time (t) it takes for the shell to hit the ground. In this equation, yi represents the initial vertical position, vi represents the initial vertical velocity, and a represents the acceleration due to gravity. The initial vertical position is 0 since the shell is launched from ground level, the initial vertical velocity is 2034.8 m/s, and the acceleration due to gravity is -9.8 m/s^2 (since it is acting in the opposite direction of the initial velocity). Solving for t, we get t = 2034.8 m/s / 9.8 m/s^2 = 207.7 s.

Finally, convert the time from seconds to kilometers by multiplying 207.7 s by 0.001 km/s to get a distance of 0.2077 km. This is the distance the shell traveled horizontally before hitting the ground.

To find the total distance the shell traveled, we can use the equation Vf = Vi + a(tf - ti) and solve for the final horizontal velocity (Vf). The final horizontal velocity would be 1095.9 m/s (since there is no horizontal acceleration) and the final time (tf) would be the same as the time we calculated previously. Solving for the initial time (ti), we get ti = tf - Vf/a = 207.7 s - (1095.9 m/s / 9.8 m/s^2) = 95.2 s.

Now, we can use the equation x = xi + vit + 1/2 at