Two balls are thrown from the top of a building a distance H from the ground. One ball is thrown vertically up with initial velocity V, while the other is thrown vertically down with velocity V. a. How long does it take the ball thrown upward to get to its highest point?
b. How high does it rise?
c. What is the difference in the time of flights for the two balls?
(Let H = 3 m, V = 24 m/s)

The ball thrown upward has height y and velocity v at time t given by

y₁ = 3 m + (24 m/s) t - 1/2 g t ²

v₁ = 24 m/s - g t

while the ball thrown downward has height and velocity

y₂ = 3 m - (24 m/s) t - 1/2 g t ²

v₂ = - 24 m/s - g t

where g = 9.80 m/s².

(a) At its highest point, the first ball has zero velocity:

24 m/s - g t = 0   →   t = (24 m/s)/g ≈ 2.45 s

(b) The first ball reaches a maximum height y [max] such that

0² - (24 m/s)² = -2 g y [max]   →   y [max] = (24 m/s)²/(2g) ≈ 29.4 m

(c) Solve for t when y₁ = 0 and y₂ = 0, then take the absolute difference between them:

0 = 3 m + (24 m/s) t - 1/2 g t ²   →   t ≈ 5.02 s

0 = 3 m - (24 m/s) t - 1/2 g t ²   →   t ≈ 0.123 s

→   |5.02 s - 0.123 s| ≈ 5.14 s

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000