Answer:
The height from ground at which they pass each other is 22.656 m
The time at which they pass each other is 1.6 sec.
Explanation:
For ball A, we have:
height = (h)a
Initial Velocity = Vi = 7 m/s
g = - 9.8 m/s² (for upward motion)
Using 2nd eqn. of motion:
h = Vi t + (1/2)gt²
(h)a = 7t + (1/2)(-9.8)t²
(h)a = 7t - 4.9t²
this is the height of ball A with reference as the building. Taking ground as reference, we have to add the height of building, that is, 24 m.
(h)a = 7t - 4.9t² + 24 ______ eqn (1)
For ball B, we have:
height = (h)b
Initial Velocity = Vi = 22 m/s
g = - 9.8 m/s² (for upward motion)
Using 2nd eqn. of motion:
h = Vi t + (1/2)gt²
(h)b = 22t + (1/2)(-9.8)t²
(h)b = 22t - 4.9t² ______ eqn (2)
Now, when the too balls pass each other, there height must be same.
Therefore,
(h)a = (h)b
using eqn (1) and eqn (2):
7t - 4.9t² + 24 = 22t - 4.9t²
22t - 7t = 24
t = 24/15
<u>t = 1.6 sec</u>
Now, for the height, at which they pass each other put t = 6sec in eqn (2)
Therefore,
h = (22)(1.6) - (4.9)(1.6)²
h = 35.2 - 12.544
<u>h = 22.656 m</u>
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