Two objects, m1 = 20g and m@ = 40g are thrown both at the same time with equal v = 56m/s. The mas m1 is projected at an angle x1 = 20 deg while m2 is at x2=50 deg. The position of a projectile, as a function of time, can be expressed by the position vector

Question

yttrium · Answer

$$r = (x _{o} +v _{ox}t)i + (y _{o}+v _{oy}t - 4.9t^2)j$$
If the objects are launched at a height of 15.0m above ground
a. With respect to m1, what is the position of m2 at t=2.5s?
b. What are the displacements of m1 and m2 at t = 3.0s?
c. How fast is m2 as viewed by m1 at t=3.0s? (Given that v = dr/dt)

anonymous · Answer

Use trig functions to separate the velocity into its respective components.

anonymous · Answer

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anonymous · Answer

So if $v=56$ at angle $20^\circ$ then $v_x=56\cos 20^\circ$ and $v_y=56\sin20^\circ$.

anonymous · Answer

@Yttrium Get it?

yttrium · Answer

Yeah, I reached that part. Then next?

anonymous · Answer

Well the initial y value is just the height and the initial x value is just 0.

anonymous · Answer

Where are you stuck?

yttrium · Answer

What is my Yo and Xo, then?

anonymous · Answer

height of 15.0m above ground 
this is $y_0$

just let $x_0$ be 0

anonymous · Answer

Have you gotten $r_1$ and $r_2$ yet?