An artillery shell is ﬁred at an angle of 79.9◦
above the horizontal ground with an initial
speed of1560m/s.
The acceleration of gravity is 9.8m/s2.
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 79.9◦
above the horizontal ground with an initial
speed of1560m/s.
The acceleration of gravity is 9.8m/s2.
Find the total time of ﬂight of the shell,
neglecting air resistance.
Answer in units of min

anonymous · Answer

The time to reach the maximum height is t=V(initial)sin(theta) / g . The total time of flight is just twice the time to reach the maximum height.

anonymous · Answer

t=1560*sin(79.9)/9.8 is this correct?

anonymous · Answer

Yes. Don't forget to convert to minutes after you get your answer.

anonymous · Answer

so I get 5 minutes and 13 seconds and the system is telling me I am incorrect. What am I doing wrong. Thank you for your help!

anonymous · Answer

That's correct, try 5.22 minutes because these online marking systems are really stupid.

anonymous · Answer

It worked... I have been going round and round trying to figure out what I was doing wrong. I have other questions. Do you mind if I post them? There are several.

anonymous · Answer

Sure.

anonymous · Answer

alright this is the second part of the question of what I just asked, finding the horizontal range?

anonymous · Answer

Assuming a flat surface the range is v^2sin2(theta)/g . You can derive this from the projectile motion equations.

anonymous · Answer

so I got 248.326 answer has to be in Km what did I do wrong?

anonymous · Answer

Show me your calculations.

anonymous · Answer

1560^2(2sin(79.9)/9.8

anonymous · Answer

not 2sin(79.9)  it's sin(2(79.9))

anonymous · Answer

why would it be sin(2(79.9)?

anonymous · Answer

Because d=2V^2cos(theta)sin(theta)/g which reduces to V^2sin2(theta)/g from 2 trig identities.

anonymous · Answer

alright so then new answer I got 85.75 km

anonymous · Answer

That sounds about right.

anonymous · Answer

and it was correct thanks next question is,
During a baseball game, a batter hits a popup
to a ﬁelder 95 m away.
The acceleration of gravity is 9.8m/s2.
If the ball remains in the air for 5.4 s, how
high does it rise?
Answer in units of m
what equation would be used?

anonymous · Answer

All your problems are related to projectile motion, why don't you just go here ? http://en.wikipedia.org/wiki/Projectile_motion

anonymous · Answer

alrighty ill try!