A particle is projected from a point on the ground surface at different angles keeping the speed of projection as constant, range obtained is plotted as function of the angle of projection with the horizontal ‘θ ’. Then the maximum height attained by the particle when it is projected at an angle 60°  with the vertical:  g=10m/s2

# A particle is projected from a point on the ground surface at different angles keeping the speed of projection as constant, range obtained is plotted as function of the angle of projection with the horizontal $\text{'}\theta \text{\hspace{0.17em}}\text{'}.$ Then the maximum height attained by the particle when it is projected at an angle  with the vertical:  $\left(\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^{2}\right)$

1. A

62.5 m

2. B

125 m

3. C

$50\sqrt{3}\text{\hspace{0.17em}}m$

4. D

31.25m

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### Solution:

$R=\frac{{u}^{2}\mathrm{sin}2\theta }{g}⇒at\theta ={45}^{0};$

$R={R}_{\mathrm{max}}=\frac{{u}^{2}}{g}=250m$

$⇒H=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta \text{'}}{2g}=\frac{250×{\mathrm{sin}}^{2}{30}^{o}}{2}$

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