What is the maximum height achieved by the projectile?

[pinky2468]

I have a question from my practice test that is supposed to be on our midterm, so I want to make sure I am doing it right!

A 10kg projectile is launched from the wall of a canyon at an angle of 30 degrees above the horizon with a velocity of 100m/s. The wall is 1500m above the canyon floor.
a)what is the maximum height achieved by the projectile?
b)what is the impact velocity with the canyon floor?

So, I did part a and I found the x and y components of v...vsin30 and vcos30.
I then found the time and used it it the equation y=Voyt + 1/2at^2 and I got y=128m

So, for part b which equation do I use? If I use pythagorean theorem the answer is 100m/s which is the same as the initial. Is that right? Or do I need to use a different equation?

 

[Pyrrhus]

why don't you use conservation of mechanical energy?

$$\Delta K + \Delta \Omega = 0$$

For max height, you know Vy = 0, and Vx is constant, so use that. There is gravitational potential energy. You could do it with kinematics, but why bother?

 

[pinky2468]

Because, we have not learned that yet so I can't use the equation

 

[ShadowHound]

Hmmm, big no on that.

You split the velocity into x-y components which is good, but your choice of formula for part 1 is somewhat questionable. See the question have not given you how long the bullet stays in the air, and therefore you would have 2 unknowns in d=ut + 1/2 at^2. it's still doable but take much more effort in figuring out the t.

Since you're only asked to figure out the height, just consider everyting in y direction. Try instead using v^2 = u^2 +2ad. (v==>final velocity, u ==> initial velocity, a ==> acceleration due to gravity, negative because it's trying to stop the bullet, d ==> distance) set v = 0 since the highest point is when the bullet is no longer going up. and give it a try. (you should get some 500ish answer)

for part b, use the same equation, but set u = 0 since you're calculating the velocity of bullet going from the top of the arc to the ground in the y direction. set d = the answer you have from part a + height of the wall, a = acceleration due to gravity, positive this time because it's accelerating the bullet.

after you figure out the final velocity in y direction, you can just add it with the x-direction velocity (which haven't changed) using pythagorean theorem.


Hope that helps

 

[Pyrrhus]

Well in that case, Remember Vy = 0 at max height. if you found the time at max height, then plug it back, show me how you found the time at max height, so i can see if its ok. By the way you got an initial position, Yo = 1500 m

 

[pinky2468]

Ok so I figured out the height w/o using the time,
V^2 = Vi^2 +2ay and I still get 128m
Then I used the same equation to get the final velocity of y and I set the initial velocity to 0 and I got 179m/s,

Then I used the pythagorean theorem to get the final velocity 199m/s

Does that seem right?

 

[Pyrrhus]

You're not taking into account the initial Y.

 

[pinky2468]

I used the same Vx because it stays constant right? Vcos30= 87m/s

 

[Pyrrhus]

Vx stays constant, yes

 

[pinky2468]

So, was my answer right or am I still doing something wrong inthe final equation?

 

[ShadowHound]

Ok so I figured out the height w/o using the time,
V^2 = Vi^2 +2ay and I still get 128m


Does that seem right?

yeah, that's about right, I forgot the Sin whe I did my sketch.

 

[pinky2468]

I did take into account the initial y... I added it to the y I found in part a for the equation to find the final velocity of y