King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 63 m/s and an angle of 42°. A cannonball that was accidentally dropped hits the moat below in 1.1 s.
How far from the castle wall does the cannonball hit the ground? which I found to be 409.23m
What is the ball's maximum height above the ground?
I thought I could use
ymax=(vosintheta)tmax-1/2g(tmax)^2 with a tmax of 4.30s which I found by using Vy=VoSintheta-gtmax=O and I obtained 90.66m for my ymax. But that came out to be wrong so I thought that I had to calculate the height of the castle wall but everything that I do come out to be wrong so idk... Please help!
3 answers
I just calculated the height of the wall to be 5.93m?
v=vosinTheta-9.8 t
at the top, vvertical is zero, so
0=63*sin43-9.8 tmax so tmax can be found when at the top.
Now, hi can be detrmined form the dropped ball.
hf=0=hi-4.9t^2
you know as 1.1, so find hi.
Now,
hmax=vo*sin43*tmax-4.9tmax^2+hi
that will do it.
at the top, vvertical is zero, so
0=63*sin43-9.8 tmax so tmax can be found when at the top.
Now, hi can be detrmined form the dropped ball.
hf=0=hi-4.9t^2
you know as 1.1, so find hi.
Now,
hmax=vo*sin43*tmax-4.9tmax^2+hi
that will do it.
Thank you! I got it! The answer is 96.59m