# Finding velocity components right before a projectile hits the ground at a point s.

I have a problem where a projectile is shot from the edge of a building at an angle and lands at a point s. After completing the first few parts and getting to the 4th, I'm now wondering about how the answer came out. That part of the question was "in the instance right before the the projectile hits the ground at point s, what are the horizontal and vertical components of the velocity?"

I have the Vx should be the same as it was initially, so nothing new to find there, but for Vy...

having found $Vy = 40 m/s$ using $Vy_0 = Vsin(\theta)$ with $\theta$ being $35$ degrees, and initial speed being 65m/s, I have:

$Vy = Vy_0 + at= 37 m/s+(-9.8m/s^2 * 9.9s)= -60m/s$

the 9.9 sec being the time it took to hit the ground.

Should the answer for this be coming out negative?

• You are measuring $y$ (positive) upwards, since acceleration is negative. So the negative velocity is downwards. – Thumbnail Feb 6 '17 at 2:11
• Where does the $9.9 \ sec$ comes from if the projectile takes $8 \ sec$ to hit the ground? And please supply the angle and initial velocity of the projectile. – Thumbnail Feb 6 '17 at 2:12
• Sorry, made a mistake, fixed it in the equation with an edit but forgot the second place I wrote 8. Supposed to be 9.9 in both. – windy401 Feb 6 '17 at 2:19