Ared 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 23.8 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/s^2. a)what is the maximum height the blue ball reaches?
b)what is the height of the blue ball 1.8 seconds after the red ball is thrown?
c)how long after the red ball is thrown are the two balls in the air at the same height?

A) The maximum height of the blue ball is 29.7 m above the ground.

B) The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) The balls are at the same height 1.41 s after the red ball is thrown

Explanation:

A) At maximum height, the velocity of the blue ball is 0 because for that instant, the ball does not go up nor down.

The equation for velocity for an accelerated object moving in a straight line is:

v = v0 + a*t

where

v = velocity.

v0 = initial velocity.

a = acceleration, in this case, it is the acceleration of gravity, g, 9.81 m/s².

t = time.

Then:

0 = v0 + g * t  (if the origin of the reference system is the ground, then g is negative)

0 = 23.8 m/s - 9.81 m/s² * t

-23.8 m/s / -9.81 m/s² = t

t = 2.43 s

With this time, we can calculate the position of the blue ball. The equation for position is:

y = y0 + v0 * t + 1/2 * g * t²

y = 0.8 m + 23.8 m/s * 2.43 s - 1/2 * 9.81 m/s² * (2.43 s)²

y = 29.7 m

the maximum height of the blue ball is 29.7 m above the ground.

B) 1.8 s after throwing the red ball, the blue ball was in the air for (1.8 s - 0.6) 1.2 s. Then, using the equation for the position of the blue ball:

y = 0.8 m + 23.8 m/s * 1.2 s - 1/2 * 9.81 m/s² * (1.2 s)² = 22.3 m

The height of the blue ball after 1.8 s of throwing the red ball is 22.3 m

C) Now, we have to find the time at which both positions are equal. Notice that the time of the blue ball is not the same as the time for the red ball. The time for the blue ball is the time of the red ball minus 0.6 s:

t blue = t red - 0.6 s

Then:

position red ball = position blue ball

 y0 + v0 * t + 1/2 * g * t² = y0 + v0 * (t- 0.6) + 1/2 * g * (t-0.6s)²

25 m + 1.2 m/s * t -1/2 * 9.81 m/s² * t² = 0.8 m + 23.8 m/s * (t-0.6 s) - 1/2 * 9.81 m/s² * (t-0.6 s)²

24.2 m + 1.2 m/s * t -4.91 m/s² * t² = 23.8 m/s * t - 14.28 m - 4.91 m/s² * (t-0.6 s)²

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² (t² - 1.2 s * t + 0.36 s²)

38.5 m - 22.6 m/s * t - 4.91 m/s² * t² = -4.91 m/s² * t² + 5.89 m/s * t - 1.77 m

40.3 m - 28.5 m/s * t = 0

t = -40.3 m / -28.5 m/s

t = 1.41 s

The balls are at the same height 1.41 s after the red ball is thrown and 0.81 s after the blue ball is thrown.

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it