A Projectile motion problem, I .

[a.ratnaparkhi]

Homework Statement

A cannon fires successively two shells with velocity u=250 m/s;the first at angle $$\theta1$$=$$\pi/3$$ & the second shell at an angle $$\theta2$$=$$\pi/4$$ to the horizontal, the azimuth being same. Neglecting air drag, find the time interval between firings leading to colling of shells.

Homework Equations

range R=(u^2sin2$$\theta$$)/g
Time of flight t=(2usin$$\theta$$)/g

The Attempt at a Solution

I'm really confused &unable to figure out.
Book says that its 11s.

 

[paragchitnis]

Find the horizontal distance where the two bodies collide by solving there equations. From this, you can get time of flight of two bodies from projection to collision and what you need..?

 

[a.ratnaparkhi]

I thought, if we assume the required time interval 't' and time of flight of two shells T1& T2 respectively, then after equating T1+t=T2, 't' can be easily found.

 

[zgozvrm]

I thought, if we assume the required time interval 't' and time of flight of two shells T1& T2 respectively, then after equating T1+t=T2, 't' can be easily found.

No...that would be the time between firings to get the shells to hit the ground simultaneously.

 

[rwisz]

I understand your confusion and will try to provide some clarity on this problem. The key concept here is projectile motion, which involves the motion of an object in a curved path due to the combined effects of its initial velocity and the force of gravity. In this case, the two shells fired from the cannon are experiencing the same initial velocity and the same gravitational force, but they are being fired at different angles.

To find the time interval between firings, we first need to find the time of flight for each shell. The time of flight equation you provided is correct, so we can use it to find the time of flight for the first shell fired at an angle of π/3. Plugging in the given values, we get t1 = (2(250 m/s)sin(π/3))/9.8 m/s^2 = 11.4 seconds.

Similarly, we can use the time of flight equation to find the time of flight for the second shell fired at an angle of π/4. Plugging in the given values, we get t2 = (2(250 m/s)sin(π/4))/9.8 m/s^2 = 10.1 seconds.

Now, since we want the shells to collide, we need to find the time interval between the firings. This can be found by subtracting the time of flight for the second shell from the time of flight for the first shell. So, the time interval between firings is 11.4 seconds - 10.1 seconds = 1.3 seconds. Therefore, the shells need to be fired 1.3 seconds apart in order to collide.

I hope this helps to clarify the problem for you. Remember, as a scientist, it's important to carefully read the problem and understand the key concepts involved in order to come up with a solution. Keep practicing and you will become more confident in solving projectile motion problems. Good luck!