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# A ball is thrown with an initial velocity of 30 m/s at an angle of 60$^{\circ}$ above the horizontal. Find the maximum height the ball can travel to?

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When the ball is thrown with velocity $v$, the maximum height attained is given by $y_{max}= \large\frac{v_y^2}{2g}$
Here $v_y$ is the $y$ component of the initial velocity $= v_0\sin \theta = 30 \times \sin 60^{\circ} = 26\;m/s$
$\Rightarrow y_{max} = \large\frac{26^2}{2 \times 9.8}$$= 34.5m$
answered Jul 16, 2014 by
edited Aug 13, 2014

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