Thread: Projectile Motion (cannonball)

A cannonball is fired with an initial speed of 90.0 m/s at an angle 61.0 ° above the horizontal (see diagram). The cannonball strikes point A on top of a cliff 10.0 s seconds after being fired. Ignore air resistance in this problem.

Calculate the speed of the cannonball just before it hits A.



My Attempt:

The things I have calculated previously are:

· Point A is 436 meters horizontally from the cannon.
· The maximum height the cannonball reaches is 316.
· The cliff is 297 m high.

Now in order to calculate the speed of the ball before it hits the A, I tried to use the formula but I do not end up with the correct answer (which is 47.7). I heard that this formula only works for finding the vertical component of the speed of the cannonball. So what do I need to do??

I even tried using to find the vertical speed and v=xt for the horizontal speed and then add up the two speeds to find the resultant. But this didn't work either...

Are you clear on how to use vectors and components of trajectories? Whenever faced with a problem containing a vector, see if you can divide it into components pointing along the coordinate axis of a coordinate system that you are familiar with. In this case a cartesian is the given choice.

Once the components have been identified, answer these questions:
What vertical and horizontal speeds does the cannon ball have at the top of the trajectory?
At what time does it reach this point?
What horizontal and vertical speeds does it have when it hits the cliff?
How do you calculate the resultant speed from its components?

Originally Posted by demode

A cannonball is fired with an initial speed of 90.0 m/s at an angle 61.0 ° above the horizontal (see diagram). The cannonball strikes point A on top of a cliff 10.0 s seconds after being fired. Ignore air resistance in this problem.

Calculate the speed of the cannonball just before it hits A.



My Attempt:

The things I have calculated previously are:

· Point A is 436 meters horizontally from the cannon.
· The maximum height the cannonball reaches is 316.
· The cliff is 297 m high.

Now in order to calculate the speed of the ball before it hits the A, I tried to use the formula but I do not end up with the correct answer (which is 47.7). I heard that this formula only works for finding the vertical component of the speed of the cannonball. So what do I need to do??

I even tried using to find the vertical speed and v=xt for the horizontal speed and then add up the two speeds to find the resultant. But this didn't work either...

v(horizontal) = the same throughout the projection = 90cos61 m/s
initial v(vertical) = 90sin61 m/s up to the max height.

using v=u+at and initial v(vertical) ,
find the time to reach max height.

0=90sin61-9.81t

I believe t = 8.02s.

then we know it took 10s, from the beginning.

so 10-8.02,
t=1.9760s.

then again using v=u+at, we can find the final velocity(vertical)

v=0+9.81*1.9760
=19.38m/s

then using
we can find the speed of the cannonball just before it hits A. required.

so just do
you have your answer = 47.74m/s.

Ah, well, call me conservative, but I don't quite see how it's a help to anyone to just present a finished solution straight up. Each to his own, I suppose.

Originally Posted by Sputnik

Ah, well, call me conservative, but I don't quite see how it's a help to anyone to just present a finished solution straight up. Each to his own, I suppose.

What I've posted is not my "homework", I already knew the solution. I just didn't know the exact working, which is what he showed me step by step. His post could not have been more helpful.

What he should of said was to use vectors. The horizontal component is figured out by the angle and the initial speed and the vertical component he already knew how to work out.