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Physics · MOTION IN A STRAIGHT LINE
हिन्दीA ball is thrown vertically upwards from the top of a building of height \( 25 \text{ m} \) with an initial speed of \( 20 \text{ m/s} \). At the exact same instant, a second ball is dropped from rest from a point \( 45 \text{ m} \) directly above the ground. Taking \( g = 10 \text{ m/s}^2 \), find the time elapsed before the two balls cross each other in air.
Correct answer · Option 4
\( 1.0 \text{ s} \)
Explanation
Taking ground as origin, height of ball 1 is \( y_1(t) = 25 + 20t - 5t^2 \) and height of ball 2 is \( y_2(t) = 45 - 5t^2 \). Setting \( y_1(t) = y_2(t) \implies 25 + 20t - 5t^2 = 45 - 5t^2 \implies 20t = 20 \implies t = 1.0 \text{ s} \). At \( t = 1.0 \text{ s} \), \( y = 40 \text{ m} \) (above the building top), so they cross in air.
