Physics help? I just cant get b).

When baseball players throw the ball in from the outfield, they usually allow it to take one bounce before it reaches the infield, on the theory the ball arrives sooner that way. Suppose the angle at which a bounced ball leaves the ground is the same as the angle at which the outfielder threw it, as in the figure, but that the ball's speed after the bounce is one half of what it was before the bounce.

(a) Assuming the ball is always thrown with the same initial speed, at what angle θ should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 35.0° with no bounce (green path)?
the answer is 24.37° i just need help with b.

(b) Determine the ratio of the times for the one-bounce and no-bounce throws.

for part b)

for projectiles that land at the same level they were thrown, their time of flight is

t= 2v0sin(theta)/g

so for the one bounce case, we have tb = 2v0 sin(thetab)/g + v0 sin(thetab)/g

the first term on the right is the time for the throw from the outfield, the second term is the time for the ball after it bounced (the velocity after the bounce is half the initial, which is way there is no coefficient of 2 in the second term)

tb is the total time for one bounce, v0 is the initial velocity, thetab is 24.37 deg,

therefore, the total time for the one bounce throw is 3v0 sin(24.37)/g

the time for the no bounce throw is tno = 2 v0 sin 35/g

now, let's take the ratio of tno (no bounces) to tb (one bounce):

tn0/tb = 2 v0 sin 35/g /3v0 sin 24.37 /g = 2 sin 35/3 sin 24.37 = 0.93

so the time for the no bounce throw is 0.93 of the one bounce throw

I would assume that you need to calculate the time taken in each scenario, given the speed (m/s) and distance (m). The ratio is the bigger one divided by the smaller.