- #1

eric_kevin

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## Homework Statement

A canon is set at an angle of 45 degrees above the horizontal. A cannonball leaves the muzzle with a speed of 2.2 x 10^2 m/s. Air resistance is negligible. Determine the cannonball's

A)Maximum height

B)Time of Flight *** this is the one I am stuck on

C)Horizontal Range (to the same vertical level)

D)Velocity at Impact

The answers are: A) 1.2 x 10^3 m

B) 32 s

C) 4.9 x 10^3 m

D) 2.2 x 10^2 m

## Homework Equations

vertical equations

v2y^2 = v1y^2 +2ay[tex]\DeltaD[/tex]

Dy=viy*t+1/2at^2

Dy=vfy*t - 1/2at^2

horizontal equations

vix =dx/t

## The Attempt at a Solution

A) vi y = (sin 45)(220m/s)

= 155.56 m/s

vf^2 = vi^2 +2ad

**rearrange to solve for d

vf^2 - vi^2 / 2a= d

0 - (155.56m/s)^2 / -19.6

= 1234.64 m

= 1.2 x 10^3 m

** so finding the maximum height isn't the problem ^

i can't seem to find the time

## Homework Statement

vi x = (cos 45)(220m/s) vi y = (sin45)(220m/s)

= 155.56 m/s = 155.56m/s

t= ? a = -9.8 m/s/s

d=? vf y (at peak) = 0 m/s