I fire a tennis ball from an air cannon, straight up from ground level, at initial speed v0. At

the instant it reaches the top of its trajectory, I fire a second tennis ball at the same initial
speed. Air resistance is negligible. At what fraction of the maximum height reached by the
first ball do the two collide?

start ball one at max height:

hf=maxheight-1/2 g t^2
ball two:
hf=vo*t-4.9t^2

but t is the same, and hf is the same:

set the two equations equal, and solve for t.
then having t, find hf.

To find the fraction of the maximum height reached by the first ball at which the two balls collide, we need to analyze the motion of the two balls.

The motion of the first ball can be described using the equations of motion for projectile motion. The maximum height reached by the first ball can be found using the equation:

H = (v₀^2) / (2g)

where H is the maximum height, v₀ is the initial speed, and g is the acceleration due to gravity.

Now, let's determine the time it takes for the first ball to reach the top of its trajectory. The time of flight for a projectile is given by:

t = v₀ / g

At this point, the second ball is fired with the same initial speed v₀. Since the second ball is fired at the same initial speed, its trajectory will be symmetrical to the first ball's trajectory. As a result, the time it takes for the second ball to reach the height at which the two balls collide is also equal to the time of flight of the first ball (t).

Now, the time it takes for the second ball to reach the height at which the two balls collide can be found using the equation:

y = v₀t - (1/2)gt^2

where y is the height at which the two balls collide. Rearranging the equation, we get:

y = t(v₀ - (1/2)gt)

Since we know that this height is equal to the fraction of the maximum height reached by the first ball, we can set y equal to the fraction of the maximum height, and solve for the fraction:

Fraction = y / H = t(v₀ - (1/2)gt) / (v₀^2 / (2g))

Simplifying, we get:

Fraction = (v₀t - (1/2)gt^2) / (v₀^2 / g)

Since t = v₀ / g, we substitute this value:

Fraction = (v₀(v₀/g) - (1/2)g(v₀/g)^2) / (v₀^2 / g)

Fraction = (v₀^2 / g - (1/2)(v₀^2 / g)) / (v₀^2 / g)

Fraction = (1/2) / 1

Fraction = 1/2

Therefore, the fraction of the maximum height reached by the first ball at which the two balls collide is 1/2.