Question

# A stone is thrown vertically at a speed of $$30 ms^{-1}$$ making an angle of $$45^o$$ with the horizontal. What is the maximum height reached by the stone ? Take $$g = 10 ms^{-2}$$.

A
30 m
B
22.5 m
C
15 m
D
10 m

Solution

## The correct option is A 22.5 mGiven :     $$u =30$$  $$m/s$$         $$\theta =45^o$$ Maximum height reached     $$H = \dfrac{u^2\sin^2 \theta}{2g}$$               $$(\sin^245^o = 0.5)$$$$\therefore$$    $$H = \dfrac{30^2\times 0.5}{2\times 10} = 22.5$$ mPhysics

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