Answer to Question #91358 in Mechanics | Relativity for Chinmaya S.C

Question

A projectile like a stone is thrown with an initial speed 10 m/s at an angle of 30 degree to the surface of a hill inclined at an angle of 20 degree to the horizontal.
If the projectiles path is imaginaryly extended, it intersects a plane parallel to the surface of the incline at an angle 90 degree to it.Then the distance between the incline and the imaginary plane is?

Answer 1

At point A (the point of the beginning of the movement) the angle between the direction of the throw and the horizontal is 30 - 20 = 10 degrees. At the imaginary point B, the angle between the direction of movement and the horizontal is 90 - 20 = 70 degrees. The task can be rephrased: the stone is thrown at an angle of 70 degrees to the horizontal, flying x meters horizontally, the stone has a speed of 10 m / s and the angle between the direction of movement and the horizontal is 10 degrees. To find the value of x, we write the equation of motion for the stone (let the origin be at point B):

$$x(t)=x_B+v_{Bx}t=v_{Bx}t \space \space (1a)$$ $$y(t)=y_B+v_{By}t-\frac {1} {2} gt^2=v_{By}t-\frac {1} {2} gt^2 \space \space (1b)$$

From (1a) we find the speed at point B:

$$v_{Ax}=v_A \cdot cos(\beta-\alpha)=10\cdot cos(\frac {\pi} {18})=\frac {dx(t)}{dt}=v_{Bx}=v_B\cdot cos(\frac {7\pi} {18})$$ $$v_B=10 \frac {cos(\pi/18)} {cos(7\pi/18)}=28.8 \space \frac m s \space \space (2)$$

From (1b) and (2) we find the time of movement (imaginary movement):

$$v_{Ay}=v_A \cdot sin(\beta-\alpha)=\frac {dy(t)}{dt}=v_{By}-\frac {g\cdot t_{AB}}{2}=v_B \cdot sin(\frac {7\pi} {18})-\frac {g\cdot t_{AB}}{2}$$ $$t_{AB}=\frac {2}{g}(v_B\cdot sin(\frac {7\pi} {18})-10\cdot sin(\frac {\pi} {18}))=5.2 \space s \space \space (3)$$

From (1a), (2) and (3):

$$x=v_{Bx} \cdot t_{AB}=28.8 \cdot 5.2 \cdot cos(7 \pi /18)=51.2 \space m$$

Answer: 51.2 m