A particle is projected at an angle of 37∘ with an inclined plane as shown in

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A particle is projected at an angle of

37^{\circ}

with an inclined plane as shown in figure. Calculate:
(i) Time of flight of particle.
(ii) Distance traveled by particle (AB) along the inclined plane

A Particle Is Projected At An Angle Of 37∘ With An Incline...

Options:

A . Time of flight = 125sec Distance along incline = 125(8−6√3)m

B . Time of flight = 2 sec Distance along incline = 16 m

C . Time of flight = 3 sec Distance along incline = 6 m

D . Time of flight = 125sec Distance along incline = (8−6√3)m

Answer: Option A
:
A
(i) To find out time of flight here, we can analyze the motion in y-direction; we can use the formula

y = u_{y} t + \frac{1}{2} a_{y} t^{2}

. By analyzing motion in y-direction , the displacement of the particle in y-direction during motion is zero.

A Particle Is Projected At An Angle Of 37∘ With An Incline...

Now

u_{y} = u s i n α = u . s i n 37^{\circ} = \frac{3}{5} \times 10 = 6 \frac{m}{s}

a_{y} = - g c o s θ = - g c o s 60^{\circ} = - 10 \times \frac{1}{2} = - 5 \frac{m}{s^{2}}

So,

y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow 0 = 6 t - \frac{5}{2} t^{2} \Rightarrow t = \frac{12}{5} s

(ii) To find out the distance traveled along AB, we have to analyze the motion in x-direction. So we have to use the formula

x = u_{x} t + \frac{1}{2} a_{x} t^{2}

Here

u_{x} = u c o s α = 10 c o s 37^{\circ} = 10 \times \frac{4}{5} = 8 \frac{m}{s} a_{x} = - g s i n θ = - 10 s i n 60^{\circ} = - 10 \times \frac{\sqrt{3}}{2} = - 5 \sqrt{3} \frac{m}{s^{2}}

And

t = \frac{12}{5} s, x = 8 \times \frac{12}{5} - \frac{1}{2} .5 \sqrt{3} {(\frac{12}{5})}^{2} = \frac{96}{5} - \frac{5 \sqrt{3}}{2} \times \frac{144}{25} = \frac{96}{5} - \frac{72 \sqrt{3}}{5} = \frac{12}{5} (8 - 6 \sqrt{3}) m

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