A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting th...
Question 1 Report
A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms\(^{-2}\)]
Answer Details
Given: horizontal velocity \(u = 15\ \text{m s}^{-1}\), height \(h = 20\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
The horizontal and vertical motions are independent. First find the time of flight from the vertical motion (initial vertical velocity is zero):
\[ h = \tfrac{1}{2}gt^2 \Rightarrow 20 = \tfrac{1}{2}\times10\times t^2 = 5t^2 \] \[ t^2 = 4 \Rightarrow t = 2\ \text{s} \]
The horizontal velocity stays constant, so the horizontal distance (range) is:
\[ R = u \times t = 15 \times 2 = 30\ \text{m} \]
The particle covers a horizontal distance of 30 m before hitting the ground.
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