Projectile Motion with an Angle?

Question

A cannon is fired over level ground at an angle of 30 degrees to the horizontal. The initial velocity of the cannonball is 400 m/s, but because the cannon is fired at an angle, the vertical component of the velocity is 200 m/s and the horizontal component is 346 m/s.

a) How long is the cannonball in the air? (Use g= 10 m/s^2 and the fact that the total time of flight is twice the time required to reach the high point.)

b) How far does the cannonball travel horizontally?

c) Repeat these calculations, assuming that the cannon was fired at a 60 degree angle to the horizontal, resulting in a vertical component of velocity of 346 m/s and a horizontal component of 200 m/s. How does the distance traveled compare to the earlier result?

I feel like I know how to do this, but I am constantly not sure. To find time for part a, do you use the formula t=square root of 2 times the height divided by the gravitational acceleration?

Answer 1

Ok, whenever you see an angle, the first thing that means is VECTORS!!!. Always remember when you are doing vector problems, handle X direction calculations and Y direction calculations INDEPENDENTLY. The only thing that will relate X and Y is the time. And remember, the time is TOTALLY dependent on Y, because as soon as an object hits the ground, it's not moving anymore.

Whoever asked the question saved you the hassle of doing the trigonometry. So, like I said, lets see how long the ball is in the air by calculating t. Look what I said about t, it is ONLY DEPENDENT ON Y!!!

So, we know that Vfy^2 = Voy^2 + 2at, and he gave us Voy, which is 200 m/s. Logic tells us that if a ball is shot up from the ground at 200 m/s, that the velocity when it gets to its highest point will be 0 (this is Vfy). So lets call up positive and down negative. We also know a=g=10. So we have:

0 = (200)^2 + 2(-10)(t)
-40,000 = -20t t = 2000 s

Remember, this t is the time it takes for the ball to get to its highest point, so laws of physics tell us that it will take the exact same time to come back down as it did to get up, so the total fliht time is t=4000 seconds

So, for part b, we can use the t we calculated to find out the horizontal distance traveled. Again, we don't need to do the trigonometry because it's done for you.

Use the formula X = Voxt + 1/2 at^2, we have Vox and t, and a is 0 because the ball is not accelerating in the X direction.

So, X = 346(4000) = 1,384,600 meters or 1.38 x 10^6 meters

You try it again with part c. Also, as practice, see if you can figure out how high the ball will go! (hint, use Y = Voyt+1/2gt^2, but be sure your signs are all consistent).

Answer 2

What you want to do is come up with a parabola that traces the path of the cannonball (specifically its height). It should be a function of time t.

-4.9t^2 + 400 * sin(30 degrees) * t

Now, set it to 0 and solve for t to figure out your times.

To determine the horizontal travel distance, take 400 * cos(30 degrees) times the length of time.

Answer 3

ninety tiers because there is extra vertical stress upward. if there is horizontal stress, that's chop up. human being above me is faulty. 40 5 tiers purely has the furthest displacement/distance. if air resistance is present day, distance traveled is shorter because air molecules are blocking off the item from vacationing further. once you're speaking about the vertical drop speed, that's a similar because the purely downward stress is gravity which continually speeds up in direction of earth at 9.82m/s.

Answer 4

let angle with horizontal be x

Time in air= t= 2*u*sinx/g

Horizontal distance travelled= R = u*u*sin2x/g

u = initial velocity, g =acc due to gravity
(Substitute and find answers.)