A red sphere is thrvery own dvery own via an initial rate of 1.1 m/s from a height of 28 meters over the ground. Then, 0.5 secs after the red ball is thrown, a blue sphere is thrown upward via an initial speed of 25.3 m/s, from a elevation of 0.8 meters above the ground. The pressure of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

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- What is the rate of the red sphere right before it hits the ground?

- How lengthy does it take the red sphere to reach the ground?

- What is the maximum height the blue sphere reaches?

- What is the elevation of the blue ball 1.9 seconds after the red sphere is thrown?

- How long after the red round is thrown are the 2 balls in the air at the same height?

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Added Tue, 29 Dec "15

Treebeard

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Solutions

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a)

v2 = u2 + 2gh

v2 = 1.12 + 2 * 9.8 * 28 = 550.01

v = 23.45 m/s

The rate of the red sphere ideal before it hits the ground = 23.45 m/s.

b)

v = u + gt

23.45 = 1.1 + 9.8 * t

t = 2.28 seconds

The time it take the red round to reach the ground = 2.28 seconds,

c)

Height above the ground = 0.8 m

v2 = u2 - 2gh

0 = 25.32 - 2* 9.8* h

h = 32.66 m.

The maximum height the blue sphere reaches = (h + 0.8) = (32.66 + 0.8) = 33.46 m.

d)

Time of travel of the blue ball = (1.9 - 0.5) = 1.4 secs.

s = ut - 0.5 gt2

s = 25.3* 1.4 - 0.5 * 9.8 * 1.42= 25.816 m.

The elevation of the blue round 1.9 secs after the red sphere is thrown = (s + 0.8) = (25.816 + .8) = 26.616 m.

e)

Let the time of take a trip of the red sphere be t seconds.

So the moment of take a trip of the blue sphere = (t - 0.5) secs.

Both the balls are at the very same elevation :

28 - (s) = 0.8 + (h) ................"s" & "h" are the displacements of the red & the blue round respectively.

28 - (ut + 0.5 gt2) = 0.8 + (ut - 0.5gt2)

28 - (1.1 t + 0.5 * 9.8 t2) = 0.8 + (25.3 (t-0.5) - 0.5*9.8*(t-0.5)2)

Now we need to settle the above equation to uncover the time after which both the balls are at the exact same height.

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28 - 1.1t - 4.9t2 = 0.8 + 25.3t - 12.65 - 4.9t2 + 4.9 t - 1.225

(28 - 0.8 + 12.65 +1.225) = (25.3 + 4.9 + 1.1) * t

t = 41.075 / 31.3 = 1.3123 = 1.31 secs (approx.)

The time after the red ball is thrown are the two balls in the air at the exact same height = 1.31 seconds.