A ball is thrown straight up from the ground with speed . At the same instant, a second ball is dropped from rest from a height , directly above the point where the first ball was thrown upward. There is no air resistance.

Question- Find the value of in terms of and so that at the instant when the balls collide, the first ball is at the highest point of its motion.

A ball is thrown straight up from the ground with speed

. 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 and g so
that at the instant when the balls collide, the first ball is at the highest
point of its motion.

To find the value of h in terms of h0 and v0, we can analyze the motions of both balls using the equations of motion. We can assume the positive direction as upwards and downwards as negative.

For the ball thrown upwards:
Initial velocity, u = v0
Final velocity, v = 0 (at the highest point)
Acceleration, a = -9.8 m/s^2 (gravitational acceleration)

Using the equation of motion:
v^2 = u^2 + 2ah
0 = v0^2 + 2(-9.8)h1 (where h1 represents the height of the first ball at the instant of collision)

Simplifying the equation:
0 = v0^2 - 19.6h1 --- Equation 1

For the ball dropped:
Initial velocity, u = 0
Final velocity, v = ? (at the instant of collision)
Acceleration, a = -9.8 m/s^2

Using the equation of motion:
v^2 = u^2 + 2ah
v^2 = 0 + 2(-9.8)h2 (where h2 represents the height of the second ball at the instant of collision)

Simplifying the equation:
v^2 = -19.6h2 --- Equation 2

At the instant the balls collide, the position of both balls must be the same. Therefore, we can equate h1 and h2.

h1 = h2

Substituting h1 for h2 in equation 2:
v^2 = -19.6h1

Substituting this value of v^2 in equation 1:
0 = v0^2 - 19.6h1

Rearranging the equation:
h1 = v0^2 / 19.6

Therefore, the value of h in terms of h0 and v0 is h1 = v0^2 / 19.6.