When `t = 10`, the particle is at `(5000, 400)`.

Here is the graph of the motion.

Note:

Graph of projectile at different times

We are told that

x = 5t3  

So

`dx/dt=15t^2`

At t = 10, the velocity in the x-direction is given by:

  `dx/dt = v_x=15(10)^2=1500\ "ms"^-1`

Also, y = 4t2   so the velocity in the y-direction is:

  `dy/dt=8t`

When t = 10, the velocity in the y-direction is:

`dy/dt=v_y=8(10)=80\ "ms"^-1`

So the magnitude of the velocity will be:

`v=sqrt((v_x)^2+(v_y)^2)=sqrt(1500^2+80^2)=1502.1\ "ms"^-1`

Now for the direction of the velocity (it is an angle, relative to the positive x-axis):

`tan\ theta_v=v_y/v_x=80/1500`

So `θ_v= 0.053` radians `= 3.05^@`.