A red ball is thrown down with an initial speed of 1.3 m/s from a height of 27 m

Question

A red ball is thrown down with an initial speed of 1.3 m/s from a height of 27 meters above the ground. Then, 0.5 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 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. What is the height of the blue ball 1.8 seconds after the red ball is thrown? How long after the red ball is thrown are the two balls in the air at the same height?

Explanation / Answer

a) 1.8 s after red ball is dropped = 1.8 - 0.5 = 1.3 s when blue ball is thrown.

s = 0. 8+ ut + (1/2)gt² = 0.8 + 24*(1.3) + (1/2)(-9.81)(1.3)² = 23.70 m

b)
Let's consider time to meet = t (after the red ball is thrown)
then red ball's distance above ground s = 27 - [1.3t + (1/2) 9.8t²] = 27 - 1.3t - 4.9t²

Now blue balls time in flight = (t - 0.5) s
then blue ball's distance = 0.8 + 24(t-0.5) -(1/2)9.8(t-0.5)²
= 0.8 + 24t - 12 - 4.9 (t² - 1.0t + 0.25)
= 24*t - 11.2 - 4.9t² + 4.9t - 1.225
= -4.9t² + 28.9t - 12.425

Since both distances are same..

27 - 1.3t - 4.9t² = -4.9t² + 28.9t - 12.425
30.2*t = 39.425
t = 39.425 / 30.2 = 1.30 s

This means 1.30s after red ball was thrown both both meet in the air

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