A ball is thrown at a speed of $20 m / s$ at an angle of $30^{o}$ with the horizontal. The maximum height reached by the ball is (use $g = 10 m / s^{2}$ )

2 m

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3 m

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4 m

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5 m

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Solution

The correct option is D $5 m$
Given : $u = 20 m / s$ $θ = 30^{o}$ $g = 10 m / s^{2}$
We get $sin θ = sin 30^{o} = \frac{1}{2}$
Maximum height reached $H = \frac{u^{2} s i n^{2} θ}{2 g} = \frac{20 \times 20 \times {(\frac{1}{2})}^{2}}{20} = 5 m$

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A ball is thrown at a speed of 20m/s at an angle of 30o with the horizontal. The maximum height reached by the ball is (use g=10m/s2)

The correct option is D 5mGiven : u=20 m/s θ=30o g=10 m/s2We get sinθ=sin30o=12Maximum height reached H=u2sin2θ2g=20×20×(12)220=5m

A ball is thrown at a speed of $20 m / s$ at an angle of $30^{o}$ with the horizontal. The maximum height reached by the ball is (use $g = 10 m / s^{2}$ )

The correct option is D $5 m$
Given : $u = 20 m / s$ $θ = 30^{o}$ $g = 10 m / s^{2}$
We get $sin θ = sin 30^{o} = \frac{1}{2}$
Maximum height reached $H = \frac{u^{2} s i n^{2} θ}{2 g} = \frac{20 \times 20 \times {(\frac{1}{2})}^{2}}{20} = 5 m$