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!
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.
Thank you! I got it! The answer is 96.59m
To determine the maximum height of the cannonball, you are on the right track using the kinematic equation:
ymax = (vosinθ)tmax - 1/2gtmax^2
where ymax represents the maximum height, vo is the initial velocity (63 m/s), θ is the launch angle (42°), tmax is the time when the cannonball reaches its maximum height, and g is the acceleration due to gravity (-9.8 m/s^2).
You correctly calculated tmax using the equation vy = vo sinθ - gtmax = 0. Rearranging the equation gives you:
tmax = vo sinθ / g
Plugging in the values gives:
tmax = 63 sin(42°) / -9.8
tmax ≈ 0.618 s
Now you can substitute this value back into the equation for ymax:
ymax = (63 sin(42°))(0.618) - 1/2(-9.8)(0.618^2)
ymax ≈ 33.33 m
So the maximum height above the ground achieved by the cannonball is approximately 33.33 m.
To determine the distance from the castle wall to where the cannonball hits the ground, you can use the equation for range:
R = v0cosθ * t
where R is the range, vo is the initial velocity (63 m/s), θ is the launch angle (42°), and t is the total time of flight.
The total time of flight can be determined by doubling the time it took for the cannonball to hit the moat (1.1 s):
t = 2 * 1.1
t = 2.2 s
Plugging in the values into the range equation:
R = (63 cos(42°)) * 2.2
R ≈ 409.23 m
So the distance from the castle wall to where the cannonball hits the ground is approximately 409.23 m, which aligns with your previous calculation.
To summarize:
- The maximum height above the ground achieved by the cannonball is approximately 33.33 m.
- The distance from the castle wall to where the cannonball hits the ground is approximately 409.23 m.