Respuesta :
The range of a projectile motion is given by:
[tex]\frac{u_o^2 sin2\theta}{g}[/tex]
where, [tex]u[/tex] is the initial speed of the projectile, [tex]\theta[/tex] is the angle of the projectile and [tex]g[/tex] is the acceleration due to gravity.
The maximum height reached is given by:
[tex]\frac{u_o^2 sin^2\theta}{2g}[/tex]
Part a
It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.
Equating the two:
[tex]\frac{u_o^2 sin2\theta}{g}=\frac{u_o^2 sin^2\theta}{2g}\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2\times 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^{-1}4=75.96^o[/tex]
Hence, the projectile was thrown at an initial angle of [tex]\theta=75.96^o[/tex].
Part b
we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.
So, we know that range is given by:
[tex] R= \frac{u_o^2sin2\theta}{g}[/tex]
It would be maximum when [tex]sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o[/tex]
Hence, [tex]R_{max}=\frac{u_o^2}{g}[/tex]
Original range,
[tex]R=\frac{u_o^2sin2\times75.96^o}{g}\\ \Rightarrow R=\frac{u_o^2}{g}\times 0.47\\ \Rightarrow R=R_{max}0.47\\ \Rightarrow R_{max}=2.125R[/tex]
Part c:
In the part a, we know that the angle of the projectile is independent of the [tex]g[/tex] i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.
