js21o3:

twinklepowderysnow:

lol

5.0s,

50.0m

50.02m/s

Is it sad that I took the time to work that out?

Yes, yes it is.

If I am correct it is more likely for him to fall for 12,25 seconds before he reaches the ground. Since 122,5m : 10,0m/s = 12,25s

His horizontal displacement would therefore also be 122,5m.

With him really hitting the ground after 245m

Which means that his final velocity is 245 : 12,25 = 20m/s

Hmm, I would have to disagree. You are using the horizontal velocity of 10m/s with the vertical displacement, and I’m guessing you’re assuming no acceleration and using s=d/t?

Velocity is not however constant and thus you have to use a suvat equation to work out the vertical component of the fall. The initial velocity is zero because Justin is being pushed horizontally - thus at first there is no downwards motion. You also know the distance and you can use acceleration as 9.81m/s/s. That means you can use s = ut + 1/2 at^2 to get the time. 

The horizontal component does have constant velocity and thus you can use s=d/t to work out horizontal displacement.

And finally you can use suvat again to find the final vertical velocity, and then draw a vector triangle with the horizontal and vertical velocities to get the total final velocity.

Haha sorry for the ramble, and sorry if that seemed rude.

I just really really like maths and physics. :B