An object is launched horizontally from a height of 45 m above the ground with a speed of 10 \text{ m/s}. The time at which the object will hit the ground is : (Take g=10 \text{ m/s}^{2})

Explanation

The horizontal speed does not affect the time of vertical free fall. Using the kinematic equation h = \frac{1}{2} g t^2, we have 45 = \frac{1}{2} (10) t^2, which simplifies to t^2 = 9, meaning t = 3 \text{ s}.

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