need help projectile motion

[F.B]

I need help on projectile motion questions because i had to miss a whole lesson for some reason and my teacher said to try some of the questions.

Anyways im stuck on this question.

 

A cannon is set at an angle of 45 degrees above the horizontal. A cannon ball leaves the muzzle with a speed of 220 m/s. Air resistance is negligible. Determine the cannonball's

a) maximum height

b) time of flight

c) horizontal range (to the same vertical level)

 

This is what i did. Im not going to do step by step on the part i got.

 

Before these i started i found VIx which is 155.6 m/s and Viy is the same

a) At max height Vfy=0( i dont get why but i know thats how you do it)

 

2ad=Vi^2 + Vf^2

d= (155.6)^2/-2(-9.8)

d= 1240m

 

b)i have no idea how to get this one i hope you can help me.

 

c)i started but i couldnt finish. This is what i did

 

d=Vit + 1/2a(t^2)

1240=155.6t - 4.9t^2

4.9t^2 - 155.6t + 1240 = 0

 

I know i need to do the quadratic formula after this but i get a negative within the radical so i dont know what to do i think i made a mistake somewhere.

 

Can you please help me?

[timo]

a) At max height Vfy=0( i dont get why but i know thats how you do it)

Because Vfy gives you the change in height. If Vfy>0 your height will be increasing within the next moment' date=' so you obviously aren´t at the peak, yet. If Vfy<0, then the height was greater just an instant before => also not the peak.

 

b)i have no idea how to get this one i hope you can help me.

The key to these kind of problems is to realize that for this part you can comepletely forget about the motion in horizontal direction. Just calculate the time it takes the cannonball to reach it´s maximum height and then to fall down.

 

c)i started but i couldnt finish. This is what i did ...

Question c) can be easily solved by the same principle as in b). Only that this time you ignore the movement in up/down-direction. Knowing the total time of flight this one becomes really easy.

 

Didn´t check your numbers in part a). The formula "2ad = Vi² + Vf²" gave me quite some headache till I understood what it was supposed to be. It´s wrong, anyways. It´s supposed to be "2ad = Vi² - Vf²". Of course, in your case that won´t matter as Vf=0.

[swansont]

Didn´t check your numbers in part a). The formula "2ad = Vi² + Vf²" gave me quite some headache till I understood what it was supposed to be. It´s wrong, anyways. It´s supposed to be "2ad = Vi² - Vf²". Of course, in your case that won´t matter as Vf=0.

 

V_f²-V_i², and ad has a dot product

[timo]

I assumed Vi = "initial velocity (at height=0)", Vf = "final velocity (at maximum height=d)", a = "gravitational constant", d = "height". Then from conservation of energy you get to Epot(h=d) + Ekin(h=d) = m*a*d + 0.5m*Vf² = 0.5m*Vi² = Ekin(h=0) => 2ad = Vi² - Vf².

 

I didn´t really look at his calculation so he might have used another convention than his nomenclature suggested to me.

 

EDIT: Oh, I think I see now what you mean swansont (negative sign because <a|d> < 0). I´m not sure how common it is to use vectors in these kind of problems.

[swansont]

EDIT: Oh, I think I see now what you mean swansont (negative sign because <a|d> < 0). I´m not sure how common it is to use vectors in these kind of problems.

 

It's a vector equation. It being not common to actually use vectors certainly explains a lot of wrong answers.