Aball is thrown straight up from the ground with speed v0. at the same instant, a second ball is dropped from rest from a height h, directly above the point where the first ball was thrown upward. there is no air resistance. (a) find the time at which the two balls collide. (b) find the value of h in terms of v0 and g such that at the instant when the balls collide, the first ball is at the highest point of its motion.

(a)

(b)

Explanation:

Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.

Using the following Newton's law of motion,

where,

= displacement

= initial velocity

= acceleration

= time

we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):

    ......(1)   (, ball is moving against gravity)

     .......(2)    (initial velocity is zero; )

Substituting from equation (1) in (2),

or,      ......(a)

(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.

This time we shall have to use another equation of motion given by,

where, = final velocity

therefore, we get for ball 1,

      ()

or,

Putting the value of in equation (2) and rearranging, we get,

which is a quadratic equation, whose solution is given by,

there are no components of a physical goal mentioned on the list of choices that you've provided.