Ablue ball is thrown upward with an initial speed of 20.6 m/s, from a height of 0.7 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6 m/s from a height of 23.9 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. how long after the blue ball is thrown are the two balls in the air at the same height?

The two balls are at the same height 3.6 s after the blue ball is thrown.

Explanation:

The height and velocity of the balls at any given time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g ·t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time "t".

First, let´s calculate the height of the blue ball when the red ball is thrown (The origin of the frame of reference is located on the ground):

y = y0 + v0 · t + 1/2 · g ·t²

y = 0.7 m + 20.6 m/s · 2.5 s - 1/2 · 9.81 m/s² · (2.5 s)²

y = 21.5 m

Now let´s calculate the velocity of the blue ball at t = 2.5 s.

v = v0 + g · t

v = 20.6 m/s - 9.81 m/s² · 2.5 s

v = -3.9 m/s

Now, let´s calculate at which time after the red ball is thrown both balls have the same height. Notice that when the red ball is thrown, the blue ball is at a height of 21.5 m and its velocity is -3.9 m/s, so its initial height will be 21.6 m and the initial velocity will be -3.9 m/s:

height blue ball = height red ball

y0 + v0 · t + 1/2 · g ·t² = y0 + v0 · t + 1/2 · g ·t²

21.5 m - 3.9 m/s · t - 1/2 · 9.81 m/s² · t² = 23.9 m - 6 m/s · t - 1/2 · 9.81 m/s² · t²

21.5 m - 3.9 m/s · t -4.9 m/s² · t² = 23.9 m - 6 m/s · t - 4.9 m/s² · t²

6 m/s · t - 3.9 m/s · t = 23.9 m - 21.5 m

2.1 m/s · t = 2.4 m

t = 2.4 m/ 2.1 m/s

t = 1.1 s

The two balls are at the same height (2.5 s + 1.1 s) 3.6 s after the blue ball is thrown.

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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