A stone is thrown vertically downwards with an initial velocity of $40 ext{ m/s}$ from the top of a building

Question

A stone is thrown vertically downwards with an initial velocity of $40 	ext{ m/s}$ from the top of a building. If it reaches the ground with a velocity of $60 	ext{ m/s}$, then the height of the building is: (take $g=10 	ext{ m/s}^2$)

Accepted Answer

1.  **Identify Variables:** 
    *   Initial velocity ($u$) = $40 	ext{ m/s}$ (downwards)
    *   Final velocity ($v$) = $60 	ext{ m/s}$ (downwards)
    *   Acceleration ($a$) = $g = 10 	ext{ m/s}^2$ (acting downwards)
    *   Displacement ($s$) = $h$ (height of the building)
2.  **Select Kinematic Equation:** Use the third equation of motion which relates velocities and displacement :
    $$v^2 - u^2 = 2as$$
3.  **Substitute Values:**
    $$(60)^2 - (40)^2 = 2(10)h$$
    $$3600 - 1600 = 20h$$
    $$2000 = 20h$$
4.  **Solve for Height ($h$):**
    $$h = \frac{2000}{20} = 100 	ext{ m}$$

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