by cosine law we know that


now using above equation



by solving above quadratic equation we have

so it is at distance 124.9 m from deer a
Answer:
V=972π
The equation I used... V=4/3π(9)^3
Differentiate the components of position to get the corresponding components of velocity :


At <em>t</em> = 5.0 s, the particle has velocity


The speed at this time is the magnitude of the velocity :

The direction of motion at this time is the angle
that the velocity vector makes with the positive <em>x</em>-axis, such that

This is Millikan's oil drop experiment.
The downward force acting on the oil drop is its weight. The upward forces are air resistance, which is negligible due to the droplet's size, and the force due to the electric field present.