A person sitting on the ground floor of a building notices through the window, of height $1.5 ext{ m}$, a ba

Question

A person sitting on the ground floor of a building notices through the window, of height $1.5 	ext{ m}$, a ball dropped from the roof of the building crosses the window in $0.1 	ext{ s}$. What is the velocity of the ball when it is at the topmost point of the window? ($g=10 	ext{ m/s}^2$)

Accepted Answer

1.  **Identify the Given Variables:**
    *   Displacement (height of the window), $s = 1.5 	ext{ m}$
    *   Time taken to cross the window, $t = 0.1 	ext{ s}$
    *   Acceleration due to gravity, $a = g = 10 	ext{ m/s}^2$
    *   Initial velocity at the top of the window, $u = ?$
2.  **Select the Kinematic Equation:** Use the second equation of motion which relates displacement, initial velocity, time, and acceleration .
    $$s = ut + \frac{1}{2}at^2$$
3.  **Substitute and Solve:**
    $$1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2$$
    $$1.5 = 0.1u + 5(0.01)$$
    $$1.5 = 0.1u + 0.05$$
    $$0.1u = 1.5 - 0.05$$
    $$0.1u = 1.45$$
    $$u = \frac{1.45}{0.1} = 14.5 	ext{ m/s}$$

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