A cannon with a muzzle speed of 1008 m/s is used to start an avalanche on a mountain slope. The target is 1900 m from the cannon horizontally and 802 m above the cannon. At what angle, above the horizontal, should the cannon be fired? (Ignore air resistance.)
Someone help please. I've tried two different approaches to this problem which were done by other students also and it worked for the same question but with different numbers. I've used this formula
x= horizontal distance of cannon from target y= meters above the cannon
g= 9.8 u= cannon's muzzle speed
r= unknown a= angle at which cannon should be fired
b= theta
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r = sqrt(x^2+y^2)
cos(b)= x/r
r*sin(2a-b)=(9.8*2^2)+y
i got 62.429 degree
and the second method used was this formula
h=(v^2sin^2theta)/2(9.8)
and i've gotten 11.643 degree
both of them are wrong, please help :(
Due tonight at 12am
nvm there was a arithmetic error
To solve this problem, we can break it down into two components: the horizontal component and the vertical component of the cannonball's motion.
Let's start with the horizontal component. The horizontal distance the cannonball travels can be found using the equation:
x = speed * time
Since the speed of the cannonball is given as 1008 m/s, and the horizontal distance is 1900 m, we can rearrange the equation to solve for time:
time = x / speed = 1900 m / 1008 m/s = 1.886 s
Now let's move on to the vertical component. We can use the equation of motion for an object in free fall:
y = u * t + 0.5 * g * t^2
where:
y = vertical distance (802 m)
u = initial vertical velocity (0 m/s)
t = time (1.886 s)
g = acceleration due to gravity (9.8 m/s^2)
Substituting the values, we can solve for the initial vertical velocity (u):
802 m = 0 * 1.886 s + 0.5 * 9.8 m/s^2 * (1.886 s)^2
Simplifying the equation, we get:
u = 802 m / (0.5 * 9.8 m/s^2 * (1.886 s)^2) ≈ 89.76 m/s
Now, let's find the magnitude of the initial velocity (v):
v = sqrt(horizontal velocity^2 + vertical velocity^2)
The horizontal velocity is the speed (1008 m/s), and the vertical velocity is the initial vertical velocity (89.76 m/s). Substituting the values:
v = sqrt((1008 m/s)^2 + (89.76 m/s)^2) ≈ 1011.97 m/s
Finally, we can find the angle by considering the vertical and horizontal velocities:
tan(a) = (vertical velocity) / (horizontal velocity)
Substituting the values:
tan(a) = 89.76 m/s / 1008 m/s
Taking the inverse tangent (arctan) of both sides:
a = arctan(89.76 m/s / 1008 m/s) ≈ 5.017 degrees
So, the angle, a, above the horizontal at which the cannon should be fired to start the avalanche is approximately 5.017 degrees.