A cannon is fired from the top of a 15m tall building at an angle of 37degrees.

Question

How far does the cannon land from the base of the building assuming it was fired at a speed of 250m/s?

Answer 1

Do your own homework dammit.

Answer 2

Let

D = distance from the base of the building where the body lands

D = R + X

where

R = horizontal distance (range) of body from the top of the building

X = horizontal distance travelled by the body as it starts descending the final 15 meters

To solve for "R", use the formula

R = Vo^2[sin (2A)]/g

where

Vo = initial velocity = 250 m/sec (given)

A = angle of launch = 37 degrees (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Substituting values,

R = (250)^2(sin 2*37)/9.8

R = 6130.50 meters

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To solve for "X" --- the analysis is follows:

At this point the body is 6130.50 meters from the top of the building and 15 meters above the ground. And, from this point, the first working equation is

Y = VoT + (1/2)(g)T^2

where

Y = 15 meters

Vo = 250(sin 37)

g = 9.8 m/sec^2

T = time for body to reach the ground

Substituting values and solving for T,

15 = 250(sin 37)T + (1/2)(9.8)T^2

The above is a quadratic equation and (after simplifying) using the quadratic formula,

T = 0.10 sec.

The second working equation for this is

X = Vo(cos 37)(T)

where all the terms have been previously defined and substituting appropriate values,

X = 250(cos 37)(0.10)

X = 19.96 meters

Therefore,  since D = R + X

D = 6130.50 + 19.96 = 6150.46 m

The body will land at 6150.46 meters away from the base of the building.

Answer 3

This is definitely a multi-step problem, I'll explain as much as I can remember from my physics days.

Draw a right triangle with the hypotenuse being 250 m/s. Find the horizontal and vertical componants of the speed by using trig. So 250sin(37) equals the right side of the triangle, which is the vertical velocity of the cannonball (how fast it's going straight up), and 250cos(37) equals the bottom of the triangle, which is the horizontal velocity of the cannonball, how fast it's going to the right.. or however you drew it.

Once you have the components, use the two velocities you have to figure out your answer. I don't remember any of my formulas, but you should have one for objects falling straight down, and one for objects in motion left to right. That's what I always screwed up on. Be sure you put in the right velocity in the right formula.

Hopefully that helped a little. Most people always had trouble figuring out the components, once you break a velocity down it's a lot easier to work with.

Answer 4

This requires both equations of motion from physics and the horizontal and vertical speed equations, i.e Vh = cos x V and Vv = sin x V.  If you calculate the total time the cannon ball is in the air first, then calculate the horizontal velocity of the ball, then use distance = time x velocity, to calculate distnce ball has travelled!