A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24 m/s, from a height of 1.2 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/s^2. 1) What is the speed of the red ball right before it hits the ground?

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

3) What is the height of the blue ball 2 seconds after the red ball is thrown?

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

Question

Tamara Jefferson

Physics

22 April, 12:49

A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24 m/s, from a height of 1.2 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/s^2. 1) What is the speed of the red ball right before it hits the ground?

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

3) What is the height of the blue ball 2 seconds after the red ball is thrown?

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

+3

Answers (1)

Know the Answer?

Schaefer · Answer 1

1. v = 22.2 m/s

2. t = 2.25 seconds

3. h = 27.05 m

4. t = 1.16 seconds

Explanation:

The questions involve motion under the influence of gravity

1. Using the formula v² = u² + 2gh

where u = 1.6 m/s; g = 9.81 m/s²; h = 25 m; v = ?

v² = (1.6) ² + 2 * 9.81 * 25

√v² = √493.06

v = 22.2 m/s

2. Using h = ut + 1/2 gt²

where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?

therefore, h = 1/2 gt²

making t subject of the formula, t = √ (2*h / g)

t = √ (2 * 25 / 9.81)

t = 2.25 seconds

3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s

using h = ut - gt²

u = 24 m/s; t = 1.6 s; g = 9.81 m/s²

note: since the ball is travelling against gravity, g is negative

h = 24 * 1.6 - 11/2 * 9.81 * 1.6²

h = 38.4 - 12.55 = 25.85 m

since height above the ground is 1.2 m,

total height h = 25.85 m + 1.2 m

h = 27.05 m

4. Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.4) seconds.

Both the balls are at the same height:

25 - s = 1.2 + h where s & h are the displacements of the red & the blue ball respectively.

25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)

25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24 (t-0.4) - 0.5*9.8 * (t-0.4) ²)

solving the equation above for the time after which both the balls are at the same height.

25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784

collecting like terms

(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t

t = 34.184 / 33.44

t = 1.16 seconds