How High Does a Projectile Go When Fired from a Cannon?

[amlovatos]

a projectile of mass of .787 kg is shot from a cannon, at height 6.8 m as shown in the figure with an initial velocity vi having a horizontal component of 5.7m/s2.

vix cos53
5.7=.6018150232
9.471348804=vi

viy=sin53 * vi

viy=.79863551*9.471348804
viy=7.564155483


Determine the MAximum height deltay the projectile achieves after leaving the end of the cannon's barrel. answer in units of m/s.

a=9.8
vi=7.564155483
vf=0

vf^2=vi^2 +2as
0=(7.564155483)^2 +2(9.8)s
57.21644817+19.6s
-57.21644817=19.65
s=2.919206539

HELP!
Find the magnitude of the velocity vector when the projectile hits the ground. Answer in units of m/s.


Find the magnitude of the angle (with respect to horizontal0 the projectile makes when I am pacting the ground. Answer in units of degrees.

 

[HallsofIvy]

a projectile of mass of .787 kg is shot from a cannon, at height 6.8 m as shown in the figure with an initial velocity vi having a horizontal component of 5.7m/s2.

vix cos53
5.7=.6018150232
9.471348804=vi

That makes little sense. You MEAN vix= vi cos(53). (You do understand that there is no figure here?)

viy=sin53 * vi

viy=.79863551*9.471348804
viy=7.564155483


Determine the MAximum height deltay the projectile achieves after leaving the end of the cannon's barrel. answer in units of m/s.

a=9.8
vi=7.564155483
vf=0

vf^2=vi^2 +2as

You know I have seen this formula posted on this forum before and wondered what it was! That's "conservation of energy" isn't it? At the top, the Kinetic energy is 0. At the bottom the kinetic energy is 1/2 m v^2. The potential energy (relative to the bottom) is mgs. Dividing by 1/2 m gives almost that formula. But I think you have the potential energy term (2as) on the wrong side! mgs will be the potential energy at the top of the trajectory as compared to the bottom. It should be on the side with the "vf" (which is 0). In other words, 2as= vi².

0=(7.564155483)^2 +2(9.8)s
57.21644817+19.6s
-57.21644817=19.65

That's -57.21644817= 19.6s, right?

s=2.919206539

Actually, that would give s= -57.../19.6. Making the change above will make it positive.

HELP!
Find the magnitude of the velocity vector when the projectile hits the ground. Answer in units of m/s.[/quote] do the same thing, basically. When the "projectile hits the ground", it will be 6.8 m lower that you "0" point so the potential energy will be lower. Basically, you will use vf²- 2a6.8= vi^2. The "-" is because the change in height is -6.8m.

Find the magnitude of the angle (with respect to horizontal0 the projectile makes when I am pacting the ground. Answer in units of degrees.

You know the magnitude of the velocity vector from the previous problem and you know that vx is still the same as vix because there is no force horizontally (until it hits the ground!). Drawing a picture you should see that v is the hypotenuse and vx the "near side" of a right triangle.