Posted by **Anonymous** on Wednesday, August 17, 2011 at 10:09pm.

Two objects are thrown simultaneously from the same height at 45° angles to the vertical with a

speed of 20 m per second; one up, the other one down. Find the difference between the heights the

objects will be at two seconds later. How are these objects moving in regards to one another?

- maths/physics -
**drwls**, Wednesday, August 17, 2011 at 10:25pm
Y1 = 20 sin45*t - (g/2)t^2

Y2 = -20 sin45*t -(g/2)t^2

Y1 - Y2 = 40 sin45*t = 56.6 m apart

V1 = 20 sin45 - gt

V2 = -20 sin45 - gt

The vertical velocity difference between them remains

V2 - V1 = -40 sin 45 = -28.3 m/s

Two seconds after release, both are going down, but V2 is going down faster.

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