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A ball is thrown up and attains a maximum height of $180 m$ . what is its initial veolocity? (Take $= 10 m / s^{2}$ )

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30 m / s

40 m / s

50 m / s

60 m / s

answer is D.

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Detailed Solution

Its initial speed is

60 m / s

.
Given data and assumptions:
Maximum height,

h = 180 m

Let the initial speed,

= u

Final velocity, after achieving maximum height,

v = 0 m / s

Acceleration due to gravity,

g = - 10 m / s^{2}

(

- ve

, as it acts downwards and ball is moving upwards)
Using 3rd equation of motion,

v^{2} - u^{2} = 2 gh

0^{2} - u^{2} = 2 (- 10) (180)

u^{2} = 3600

u = \sqrt{3600}

u = 60 m / s

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