Answer:
R = 85.73 m
Explanation:
It is given that,
The vertical speed of a rock, [tex]v_y=35\ m/s[/tex]
The horizontal speed of a rock, [tex]v_x=12\ m/s[/tex]
We know that,
The vertical speed,
[tex]v_y=v\sin\theta=35\ ....(1)[/tex]
Horizontal speed,
[tex]v_x=v\cos\theta=12\ ....(2)[/tex]
Dividing equation (1) by (2) such that,
[tex]\dfrac{v_y}{v_x}=\dfrac{v\sin\theta}{v\cos\theta}\\\\\dfrac{35}{12}=\tan \theta\\\\\theta=\tan^{-1}(\dfrac{35}{12})\\\\\theta=71.07^{\circ}[/tex]
Let v is the initial velocity of the projection. So,
[tex]v=\sqrt{v_x^2+v_y^2} \\\\v=\sqrt{12^2+35^2} \\\\v=37\ m/s[/tex]
We need to find the distance where it will land i.e. the range of the projectile. It is given by the formula as follows :
[tex]R=\dfrac{v^2\sin2\theta}{g}\\\\\text{Putting all the values, we get }\\\\R=\dfrac{37^2\times \sin(2\times 71.07)}{9.8}\\\\R=85.73\ m[/tex]
So, it will land at a distance of 85.73 m.
