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  4. <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-qooten7cphdyaumu.jpg /> A particle is projected with speed 50ms- 1 at an angle 37° with the horizontal and it touches an inclined plane of inclination 30° as shown in figure. Height of the point from the ground where particle touches the incline is (g = 10 m s- 2)

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Q. Question
A particle is projected with speed $50ms^{- 1}$ at an angle $37^\circ $ with the horizontal and it touches an inclined plane of inclination $30^\circ $ as shown in figure. Height of the point from the ground where particle touches the incline is $\left(g = 10 m s^{- 2}\right)$

NTA AbhyasNTA Abhyas 2020

Solution:

$\frac{v \sqrt{3}}{2}=50\times \frac{4}{5}\Rightarrow v=\frac{80}{\sqrt{3}}$
Now using conservation of mechanical energy $\frac{\left(80\right)^{2}}{3}-\left(50\right)^{2}=-2gh$
$\Rightarrow \frac{- 6400 + 7500}{3}=2gh$
$\Rightarrow \frac{1100}{3 \times 10 \times 2}=h\Rightarrow h=\left(\frac{55}{3}\right)m$