Question

A cannonball is shot (from ground level) with an initial horizontal velocity of 41 m/s and an initial vertical velocity of 22 m/s.



1) What is the speed of the cannonball 1.7 seconds after it was shot?


2) How high above the ground is the cannonball 1.7 seconds after it is shot?

Answer 1

1) the horizontal velocity stays constant
... the vertical velocity is decreased by gravitational acceleration ... 9.8 m/s^2
... 22 - (1.7 * 9.8)
... add the two velocity vectors to find the resultant

2) you have the vertical velocity at 1.7 s
... average it with the initial (22 m/s)
... multiply the average by 1.7 s to find the height

Answer 2

Vo = Xo + Yo = 41 + 22i m/s.

1. V = Y = Yo + g+t = 22 - 9.8*1.7 = 5.34 m/s.
Vo = Xo + Yi = 41 + 5.34i = (Convert to polar form).

2. Y^2 = Yo^2 + 2g*h.
5.34^2 = 22^2 - 19.6h, h = ?.