Two students are on a balcony 19.1 m above the street. One student throws a ball

Question

Two students are on a balcony 19.1 m above the street. One student throws a ball vertically downward at 13.8 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. What is the magnitude of the velocity of the first ball as it strikes the ground? Answer in units of m/s. What is the magnitude of the velocity of the second ball as it strikes the ground? What is the difference in the time the balls spend in the air? Answer in units of s. How far apart are the balls 0.598 s after they are thrown?

Explanation / Answer

given initial velocity u = 13.8 m/s

h = 19.1 m

1) first ball

v^2 - u^2 = 2 * a * s

v^2 - 13.8^2 = 2*(-9.8) * (-19.1)

v = 23.76 m/s

2) for second ball

when it misses the window it has same initial velocity 13.8 m/s

so when it strikes the ground

v = 23.76 m/s

3)

time for first ball is from

s = ut + (1/2 * a*t^2)

-19.1 = -13.8 t - (0.5*(9.8)*(t)^2

t = 1.01 s

for second ball

s = ut + (1/2 * a*t^2)

-19.1 = 13.8 t - (0.5*(9.8)*(t)^2

t = 3.83 s

difference is 3.83 - 1.01 = 2.82 s

4)

for first ball

s = ut + (1/2 * a*t^2)

s1 = -(13.8 * 0.598) - (0.5*(9.8)*(0.598)^2
s1 = 10 m ( down ward)

for second ball

s2 = (13.8 * 0.598) - (0.5*(9.8)*(0.598)^2

s2 = 6.5 m ( upward)

they are apart by a distance of s = 10 + 6.5 m

s = 16.5 m

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