A projectile is launched with a speed of 10 m-s at an angle 60- with the horizontal from a sloping surface of inclination 30- The range R is Take g - 10 m-s2

At B, S_y = 0 ∴ u_y t+12a_yt^2 = 0 or t = 2u_ya_y = 2(10) 10×√(3)/2 = 4√(3)s Now, AB = R = 12a_xt^2 = 12(10×12)(163) = 13.33 m. ...

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Byju's Answer

Standard XII

Physics

2D Collision Example

A projectile ...

Question

A projectile is launched with a speed of 10 m/s at an angle $60^{\circ}$ with the horizontal from a sloping surface of inclination $30^{\circ}$ . The range R is $(T a k e g = 10 m / s^{2})$

4.9 m

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13.3 m

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9.1 m

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12.6 m

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Solution

The correct option is B 13.3 m
At B, $S_{y} = 0$

$∴ u_{y} t + \frac{1}{2} a_{y} t^{2} = 0$
or $t = - \frac{2 u_{y}}{a_{y}} = \frac{- 2 (10)}{- 10 \times \sqrt{3} / 2} = \frac{4}{\sqrt{3}} s$
Now, $A B = R = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} (10 \times \frac{1}{2}) (\frac{16}{3}) = 13.33 m$ .

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A projectile is launched with a speed of 10 m/s at an angle 60∘ with the horizontal from a sloping surface of inclination 30∘. The range R is (Take g=10 m/s2)

The correct option is B 13.3 mAt B, Sy=0 ∴uy t+12ayt2=0 or t=−2uyay=−2(10)−10×√3/2=4√3s Now, AB=R=12axt2=12(10×12)(163)=13.33 m.

A projectile is launched with a speed of 10 m/s at an angle $60^{\circ}$ with the horizontal from a sloping surface of inclination $30^{\circ}$ . The range R is $(T a k e g = 10 m / s^{2})$

The correct option is B 13.3 m
At B, $S_{y} = 0$

$∴ u_{y} t + \frac{1}{2} a_{y} t^{2} = 0$
or $t = - \frac{2 u_{y}}{a_{y}} = \frac{- 2 (10)}{- 10 \times \sqrt{3} / 2} = \frac{4}{\sqrt{3}} s$
Now, $A B = R = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} (10 \times \frac{1}{2}) (\frac{16}{3}) = 13.33 m$ .