An artillery shell is fired at an angle of 34.1 degrees above the horizontal ground with an initial speed of 1960 m/s. The acceleration of gravity is 9.8 m/s^2.

Find the total time of flight of the shell, neglecting air resistance. Answer in units of min

Question

An artillery shell is fired at an angle of 34.1 degrees above the horizontal ground with an initial speed of 1960 m/s. The acceleration of gravity is 9.8 m/s^2. 

Find the total time of flight of the shell, neglecting air resistance. Answer in units of min

ybarrap · Answer

Everything depends on the vertical component of motion. Find the time by determining the y-component of velocity (1960 * sin(34.1)). So what equations are relevant here?

anonymous · Answer

I tried to use the equation [t=2vsintheta/g\] but I got 224.256? And it was wrong

ybarrap · Answer

Have you tried $v_f = at + v_0$, setting $v_f=0$?

ybarrap · Answer

I get 112.13 seconds.

anonymous · Answer

I tried it now and it says it's still incorrect :/

ybarrap · Answer

Actually, I didn't multiply times 2, and your initial answer is what I would have found

ybarrap · Answer

they are asking for mins

ybarrap · Answer

divide by 60

anonymous · Answer

Ya I have done that too and I got about 3.73, I don't understand why it is wrong... but thanks for the help :)

ybarrap · Answer

ok