Two students are on a balcony 20.7 m above the street. One student throws a ball

Question

Two students are on a balcony 20.7 m above the street. One student throws a ball (ball 1) vertically downward at 14.0 m/s; at the same instant, the other student throws a ball (ball 2) vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a) What is the difference in the two ball's time in the air?
s

(b) What is the velocity of each ball as it strikes the ground?

(c) How far apart are the balls 0.900 s after they are thrown?
m

ball 1    magnitude   m/s direction ---Select--- upward downward ball 2 magnitude   m/s direction ---Select--- upward downward

Explanation / Answer

(a) Use the equation -

s = ut + 0.5at²

For ball thrown down:

-20.7 = (-14.0)t + 0.5(-9.8)t²

4.9t² + 14t – 20.7 = 0

So, t = [-14 + sqrt(14^2 + 2*4.9*20.7)] / (2*4.9) = 2.04 s

Other value of t is rejected because time cannot be negative.

Now, for ball thrown up:

0 = 14t + 0.5(-9.8)t²

=> t = 0 (initially at ground level)
t = 2.8 s

Therefore, difference in time = 2.8 – 2.04 = 0.76 s.

(b) Since it is constant acceleration, the ball thrown up from ground level will have the same velocity when it hits the ground as the velocity it was thrown at.

for ball thrown down:

v² = u² + 2as

v² = (-14)² + 2(-9.8)(-20.7)

v² = 601.72

v = ± 24.5 m/s

take the negative value as we know the ball is going down

velocity of ball thrown up: -14.0 m/s (Downward)
velocity of ball thrown down: -24.5 m/s (Downward)

(c) s = s0 + ut + 0.5at²

Ball thrown up:

s = 0 + (14.0)(0.9) + 0.5(-9.8)(0.9)²

s = 12.60 - 3.97 = 8.63 m

Ball thrown down:

s = 20.7 + (-14.0)(0.9) + 0.5(-9.8)(0.9)²

s = 20.7 – 12.6 - 3.97 = 4.13 m

Therefore, distance apart = 8.63 – 4.13 = 4.50 m

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