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=\frac{2v_0\sin(\alpha)}g`, and it has thus traveled
`R=x(t_0)=2\frac{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=\frac{v_0\sin(\alpha)}g`. At this time it has a height of
`y(t_1)=(v_0^2\sin^2(\alpha))/(2g)`

Therefore the problem is resolved when
`2\frac{v_0^2}g\sin(\alpha)\cos(\alpha)=(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

Step-by-step explanation:

The rock thrown will follow projectile motion with

Range ,
`R=(u^(2)\sin 2\Theta )/(g)`

and maximum height ,
`H_(max)=(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)`

=>
`(u^(2)\sin 2\Theta )/(g)=(u^(2)\sin ^(2)\Theta )/(2g)=>2\sin \Theta \cos \Theta =(\sin ^(2)\Theta )/(2)`

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

Thus the rock is thrown at an angle 76 degrees.