Respuesta :
Newton's second laws tells us that the vertical motion of the rock in function of the time [tex]t[/tex] follows [tex]y(t)=-\frac12gt^2+v_0t\cdot\sin(\alpha)[/tex] while the horizontal motion follows [tex]x(t)=v_0t\cdot\cos(\alpha)[/tex] where [tex]\alpha[/tex] is the angle of the throw with respect to the ground and [tex]v_0[/tex] is the initial speed of the rock.
Let's find the horizontal range [tex]R[/tex] : the rock hits the ground when [tex]y(t)=0[/tex] which happens for [tex]t_0=\dfrac{2v_0\sin(\alpha)}g[/tex], and it has thus traveled [tex]R=x(t_0)=2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)[/tex].
Let's find the maximum height traveled : the maximum height is obtained when the rock's vertical speed is zero, i.e. when [tex]v_0\sin(\alpha)-gt=0[/tex] that is to say at [tex]t_1=\dfrac{v_0\sin(\alpha)}g[/tex]. At this time it has a height of [tex]y(t_1)=\dfrac{v_0^2\sin^2(\alpha)}{2g}[/tex]
Therefore the problem is resolved when [tex]2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)=\dfrac{v_0^2\sin^2(\alpha)}{2g}[/tex] i.e. for [tex]\boxed{\alpha=\arctan(4)\approx76^\circ}[/tex]
Let's find the horizontal range [tex]R[/tex] : the rock hits the ground when [tex]y(t)=0[/tex] which happens for [tex]t_0=\dfrac{2v_0\sin(\alpha)}g[/tex], and it has thus traveled [tex]R=x(t_0)=2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)[/tex].
Let's find the maximum height traveled : the maximum height is obtained when the rock's vertical speed is zero, i.e. when [tex]v_0\sin(\alpha)-gt=0[/tex] that is to say at [tex]t_1=\dfrac{v_0\sin(\alpha)}g[/tex]. At this time it has a height of [tex]y(t_1)=\dfrac{v_0^2\sin^2(\alpha)}{2g}[/tex]
Therefore the problem is resolved when [tex]2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)=\dfrac{v_0^2\sin^2(\alpha)}{2g}[/tex] i.e. for [tex]\boxed{\alpha=\arctan(4)\approx76^\circ}[/tex]
Answer: The rock is thrown at an angle 76 degrees
Explanation:
The rock thrown will follow projectile motion with
Range , [tex]R=\frac{u^{2}\sin 2\Theta }{g}[/tex]
and maximum height , [tex]H_{max}=\frac{u^{2}\sin ^{2}\Theta }{2g}[/tex]
where u=initial speed of rock , g= acceleration due to gravity ,
[tex]\Theta = initial\, angle \, of \, through\, with\, horizonatal[/tex]
Given [tex]R=H_{max}[/tex]
=>[tex]\frac{u^{2}\sin 2\Theta }{g}=\frac{u^{2}\sin ^{2}\Theta }{2g}=>2\sin \Theta \cos \Theta =\frac{\sin ^{2}\Theta }{2}[/tex]
=>[tex]\tan \Theta =4=>\Theta =\tan^{-1}(4)= 76^{\circ}[/tex]
Thus the rock is thrown at an angle 76 degrees.
