A projectile is thrown at a speed of 20 m-s at an angle of 53- with the horizontal- When it is at the highest point- its center of curvature is at g - 10 m-s2

Maximum height, say, is H. H = u^2 sin^2 53^∘/2g = 20^2̸/2̸0̸×4^2/5^2 ⇒ H = 12.8 m By a_c = v^2/R_c⇒ R_C = v^2/a_c = u^2cos^2θ/10= 12^2/10= 14.4 m So, it is 1. ...

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Byju's Answer

Standard XII

Physics

The Symmetry of the Path Traced

A projectile ...

Question

A projectile is thrown at a speed of 20 m/s at an angle of $53^{\circ}$ with the horizontal. When it is at the highest point, its center of curvature is at $(g = 10 m / s^{2})$

Ground

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1.6 m below ground

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2 m above ground

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1.2 m below ground

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Solution

The correct option is B 1.6 m below ground

Maximum height, say, is $H$ .
$H = \frac{u^{2} s i n^{2} 53^{\circ}}{2 g} = \frac{20^{/ 2}}{/ 20} \times \frac{4^{2}}{5^{2}} \Rightarrow H = 12.8 m$
By $a_{c} = \frac{v^{2}}{R_{c}} \Rightarrow R_{C} = \frac{v^{2}}{a_{c}}$ $= \frac{u^{2} {cos}^{2} θ}{10} = \frac{12^{2}}{10} = 14.4 m$
So, it is 1.6 m below the ground.

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A projectile is thrown at a speed of 20 m/s at an angle of 53∘ with the horizontal. When it is at the highest point, its center of curvature is at (g=10 m/s2)

The correct option is B 1.6 m below ground Maximum height, say, is H. H=u2sin253∘2g=20/2/20×4252⇒H=12.8 m By ac=v2Rc⇒RC=v2ac =u2cos2θ10=12210=14.4 m So, it is 1.6 m below the ground.

A projectile is thrown at a speed of 20 m/s at an angle of $53^{\circ}$ with the horizontal. When it is at the highest point, its center of curvature is at $(g = 10 m / s^{2})$