Answer:
The horizontal distance covered by the firework will be [tex]\frac{1876.8}{g}[/tex]
Explanation:
Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.
writing equation of motion in vertical direction:
[tex]v_{y}=u_{y}+(-g) t[/tex]
[tex]u_{y}= u\sin \phi[/tex]
and [tex]v_{y}=0[/tex]
therefore [tex]\frac{u\sin \phi }{g} =t[/tex]
writing equation of motion in horizontal direction:
[tex]s_{x}=u_{x}t[/tex]
[tex]u_{x} = u\cos \phi[/tex]
therefore the equation becomes [tex]s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}[/tex]
therefore horizontal distance traveled =[tex]\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}[/tex]
