Question

(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

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

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*0.8382

d1 = -15.67 m

displacement of second ball in 0.838 s,,

d2 = 14.6*0.838 - (1/2)*9.8*0.8382

d2 = 8.793 s

Difference in displacement,

d = d2 - d1 = 8.793 - (-15.67)

d = 24.47 m

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