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)

Question

<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) (A) (55/3)m (B) 55m (C) 40m (D) 18m

Tardigrade Admin · Accepted Answer

(v √3/2)=50× (4/5)⇒ v=(80/√3) Now using conservation of mechanical energy ((80)2/3)-(50)2=-2gh ⇒ (- 6400 + 7500/3)=2gh ⇒ (1100/3 × 10 × 2)=h⇒ h=((55/3))m

Q.
A particle is projected with speed $50 m s^{- 1}$ at an angle $3 7^{\circ}$ with the horizontal and it touches an inclined plane of inclination $3 0^{\circ}$ as shown in figure. Height of the point from the ground where particle touches the incline is $(g = 10 m s^{- 2})$

$\frac{55}{3} m$

$55 m$

$40 m$

$18 m$

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

Q. 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=10ms−2)

355​m

55m

40m

18m