Answer:
The stone will reach 3.2 m along the surface of the hill before stopping.
Explanation:
Let the height reached by the stone be 'h' m
Given:
Mass of the stone is 7.5 kg.
Angle of inclination is, [tex]\theta = 41\°[/tex]
Initial velocity of the stone at the bottom is, [tex]v=8.5\ m/s[/tex]
Final velocity of the stone at the top is 0 m/s.
Now, as per conservation of energy principle, sum of total energy is always conserved. Therefore, decrease in kinetic energy is equal to the increase in its potential energy.
Increase in Potential Energy = Decrease in Kinetic Energy
⇒ [tex] mgh = \frac{1}{2}mv^2[/tex]
⇒ [tex] gh= \frac{1}{2}v^2[/tex]
⇒ [tex]h=\frac{v^2}{2g}[/tex]
Plug in the given values and solve for 'h'. This gives,
[tex]h=\frac{8.5^2}{2\times 9.8}=3.686\ m[/tex]
Now, slant height is given using the trigonometric ratio. The slant length 'L' is the hypotenuse and 'h' is the opposite side. Therefore,
[tex]\tan \theta=\frac{L}{h}\\\\L=h\times \tan\theta\\\\L=3.686\times \tan(41)\\\\L=3.2\ m[/tex]
Therefore, the stone will reach 3.2 m along the surface of the hill before stopping.
