A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 m

Question

A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.8 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

1)

What is the speed of the red ball right before it hits the ground?

m/s

2)

How long does it take the red ball to reach the ground?

s

3)

What is the maximum height the blue ball reaches?

m

4)

What is the height of the blue ball 1.9 seconds after the red ball is thrown?

m

5)

How long after the red ball is thrown are the two balls in the air at the same height?

s

6)

Which statement is true regarding the blue ball?

After it is released and before it hits the ground, the blue ball is always moving faster than the red ball at any given time.

After it is released and before it hits the ground, the blue ball is sometimes moving faster than the red ball at any given time.

After it is released and before it hits the ground, the blue ball is never moving faster than the red ball at any given time.

7)

Which statement is true about the red ball?

The acceleration is positive and it is speeding up

The acceleration is negative and it is speeding up

The acceleration is positive and it is slowing down

The acceleration is negative and it is slowing down

Explanation / Answer

1) initial speed of red ball u = 1.2 m/s
acceleration due to gravity g = 9.81 m/s²
distance above ground s = 25 m
final velocity v = ?

v² = u² + 2gs
v = (u² + 2gs)
v= [(1.2)² + (9.81)(25)] = 15.7 m/s

2) v = u + gt
t = (v - u)/g = (15.7-1.2)/9.81 = 1.48 s

3) For blue ball
initial velocity u = 23 m/s
final velocity v = 0
s = distance above ground

g = -9.81 ( -ve because gravity is downwards and ball is moving upwards)

s = 0.8 + (v² - u²)/2g = 0.8 + [0 - (23)²]/2(-9.81) = 26.92 m


4) 1.9 s after red ball is dropped = 1.9 - 0.6 = 1.3 s when blue ball is thrown.

s = 0.8 + ut + (1/2)gt² = 0.8 + 23(1.3) + (1/2)(-9.81)(1.3)² = 22.41 m

5)
Let's consider time to meet = t (after the red ball is thrown)
then red ball's distance above ground s = 25 - [1.2t + (1/2) 9.8t²] = 25 - 1.2t - 4.9t²

Now blue balls time in flight = (t - 0.6) s
then blue ball's distance = 0.8 + 23.1(t-0.6) -(1/2)9.8(t-0.6)²
= 0.8 + 23t - 13.86 - 4.9 (t² - 1.2t + 0.36)
= 23t - 13.06 - 4.9t² + 5.88t - 1.76
= -4.9t² + 28.88t - 14.82

Since both distances are same..

25 - 1.2t - 4.9t² = -4.9t² + 28.88t - 14.82

30.18t = 39.82
t = 39.82/30.18 = 1.32 s

This means 1.31s after red ball was thrown both both meet in the air

6) After it is released and before it hits the ground, the blue ball is sometimes moving faster than the red ball at any given time

7)   The acceleration is positive and it is speeding up

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