Question

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?

Answer 1

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 2

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.