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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