The height of a mobile phone tower is 125 m. If a device is dropped from the tower, it reaches the ground in 5 s. The value of acceleration of the device just before hitting the ground
Correct answer: A. will be 10 m/s2.
- A. will be 10 m/s2.
- B. will be 25 m/s2.
- C. will be 50 m/s2
- D. cannot be determined.
Explanation
. * Verifying with Given Values: We can verify the implied value of 'g' using the given height (h = 125 m) and time (t = 5 s) with the kinematic equation: h = ut + (1/2)gt² Since the device is dropped, the initial velocity (u) is 0 m/s. 125 = (0)(5) + (1/2)g(5)² 125 = (1/2)g(25) 125 = 12.5g g = 125 / 12.5 g = 10 m/s²
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About Acceleration
Acceleration measures the rate at which velocity changes with time, including changes in speed, direction or both. Work includes average and instantaneous acceleration, the relation a = Δv/Δt, uniform acceleration equations, and interpreting velocity time and displacement time graphs, with acceleration carefully distinguished from velocity.
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