(a)
Assume upwards direction is positive and downward is negative.
Let, time taken by the first ball to reach the ground = t1
From kinematic equation,
s = ut + (1/2)at^2
-h = u1*t1 - (1/2)*gt1^2
-20.7 = -14.6*t1 - (1/2)*9.8*t1^2
4.9*t1^2 + 14.6*t1 - 20.7 = 0
By solving above quadratic equation,
t1 = 1.0487 s
Velocity of first ball when strikes the ground,
v1 = u1 + at1
v1 = -14.6 - 9.8*1.0487
v1 = -24.87 m/s
(b)
time taken by the second ball to reach the ground = t2
From kinematic equation,
s = ut + (1/2)at^2
-h = u1*t1 - (1/2)*gt1^2
-20.7 = 14.6*t1 - (1/2)*9.8*t1^2
4.9*t1^2 - 14.6*t1 - 20.7 = 0
By solving above quadratic equation,
t2 = 4.028 s
Velocity of second ball when strikes the ground,
v2 = u2 + at2
v2 = 14.6 - 9.8*4.028
v2 = -24.87 m/s
(c)
Difference in time of both the balls,
t = t2 - t1 = 4.028 - 1.0487
t = 2.97 s
(d)
displacement of first ball in 0.838 s,
d1 = -14.6*0.838 - (1/2)*9.8*0.838^{2}
d1 = -15.67 m
displacement of second ball in 0.838 s,,
d2 = 14.6*0.838 - (1/2)*9.8*0.838^{2}
d2 = 8.793 s
Difference in displacement,
d = d2 - d1 = 8.793 - (-15.67)
d = 24.47 m
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