Two students are on a balcony 23.4 m above the street. One student throws a ball, b1, vertically downward at 15.5 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a) What is the difference in time the balls spend in the air?
?? s

(b) What is the velocity of each ball as it strikes the ground?
Velocity for b1?? m/s
Velocity for b2?? m/s


(c) How far apart are the balls 0.480 s after they are thrown?
?? M

Somewhere in your text or background info you've been given a position, or height, function for the objects. It'll be of the form h(t), expressing the height of the ball at t seconds. The function will include a provision for the initial velocity of the ball, which could be positive or negative in any given situation, depending on whether the ball is initially traveling up or down. So in your situation b1 and b2 will have different initial velocity constants in the height function. Also note that h(0) is 23.4 for both.

For (a), find the zeros of the function for both b1 and b2. (They'll be quadratics, so only the positive values make sense here.) This will be the time t at which the height of each ball is zero (i.e. when it smacks the ground).

For (b) you'll need the velocity function for each, which is the first derivative of the height function. Determine the value of the velocity function at t = ? Which you found above as the time when the balls strike the ground.

For (c) it's back to the height functions. Just solve each for t = 0.48 and look at the difference in heights at that instant of time.

Can I have the answers please? I would really appreciate it, I'm just checking