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A rock is thrown upward from the level ground in such a way that the maximum height of its flight is equal to its horizontal range R.

At what angle is the rock thrown?

Verified Answer :

Hippalectryon

Newton's second laws tells us that the vertical motion of the rock in function of the time \(t\) follows \(y(t)=-\frac12gt^2+v_0t\cdot\sin(\alpha)\) while the horizontal motion follows \(x(t)=v_0t\cdot\cos(\alpha)\) where \(\alpha\) is the angle of the throw with respect to the ground and \(v_0\) is the initial speed of the rock.

Let's find the horizontal range \(R\) : the rock hits the ground when \(y(t)=0\) which happens for \(t_0=\dfrac{2v_0\sin(\alpha)}g\), and it has thus traveled \(R=x(t_0)=2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)\).

Let's find the maximum height traveled : the maximum height is obtained when the rock's vertical speed is zero, i.e. when \(v_0\sin(\alpha)-gt=0\) that is to say at \(t_1=\dfrac{v_0\sin(\alpha)}g\). At this time it has a height of \(y(t_1)=\dfrac{v_0^2\sin^2(\alpha)}{2g}\)

Therefore the problem is resolved when \(2\dfrac{v_0^2}g\sin(\alpha)\cos(\alpha)=\dfrac{v_0^2\sin^2(\alpha)}{2g}\) i.e. for \(\boxed{\alpha=\arctan(4)\approx76^\circ}\)

Answer: The rock is thrown at an angle 76 degrees

Explanation:

The rock thrown will follow projectile motion with

Range , \(R=\frac{u^{2}\sin 2\Theta }{g}\)

and maximum height , \(H_{max}=\frac{u^{2}\sin ^{2}\Theta }{2g}\)

where u=initial speed of rock , g= acceleration due to gravity ,

\(\Theta = initial\, angle \, of \, through\, with\, horizonatal\)

Given \(R=H_{max}\)

=>\(\frac{u^{2}\sin 2\Theta }{g}=\frac{u^{2}\sin ^{2}\Theta }{2g}=>2\sin \Theta \cos \Theta =\frac{\sin ^{2}\Theta }{2}\)

=>\(\tan \Theta =4=>\Theta =\tan^{-1}(4)= 76^{\circ}\)

Thus the rock is thrown at an angle 76 degrees.

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