Question

A cannon ball fired at an angle of 15o with a velocity of 290 m/s lands 38 seconds later. What is the vertical velocity of the ball at 19 seconds?

Answer 1

Step-by-step explanation:

I will take this as 15 degrees

INITIAL Vertical velocity is given by 290 * sin 15

Gravity begins to slow it down as soon as it is launched...the vertical velocity then becomes

290sin 15 - 9.81 t sub in t = 19 to find

`v_(19)` = -111.3 m/s Which shows it is coming down

I think you have some numbers incorrect

on level ground, ignoring air friction...it should spend half the time ascending and half the time descending...

at 19 seconds it would then be = 0 m/s.... perhaps this object was fired from atop a cliff??