Question

030 (part 1 of 4) 10.0 points Two students are on a balcony 20.7 m above the street. One student throws a ball ver- tically downward at 14.6 m/s. At the same instant, the other student throws a ball verti- cally 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. 031 (part 2 of 4) 10.0 points What is the magnitude of the velocity of the second ball as it strikes the ground? Answer in units of m/s. 032 (part 3 of 4) 10.0 points What is the difference in the time the balls spend in the air? Answer in units of s. 033 (part 4 of 4) 10.0 points How far apart are the balls 0.838 s after they are thrown? Answer in units of m
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Answer #1

(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,

\Deltat = t2 - t1 = 4.028 - 1.0487

\Deltat = 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,

\Deltad = d2 - d1 = 8.793 - (-15.67)

\Deltad = 24.47 m

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