Answer:
Explanation:
Given
Initial velocity(u)=48.3 m/s
Launch angle[tex]=26^{\circ}[/tex]
On earth Range[tex]=\frac{u^2sin^2\theta }{2g}[/tex]---1
g on distant planet is different thus
[tex]R'=\frac{u^2sin^2\theta }{2g'}[/tex]----2
Divide 2 & 1
[tex]\frac{R'}{R}=\frac{g}{g'}[/tex]
[tex]3=\frac{g}{g'}[/tex]
[tex]g'=\frac{g}{3}[/tex]
(a)maximum height
[tex]h_{max}=\frac{u^2sin^2\theta }{2\frac{g}{3}}[/tex]
[tex]h_{max}=\frac{48.3^2\times \sin^226\times 3}{2\times 9.8}[/tex]
[tex]h_{max}=156.53 m[/tex]
(b)Range of ball
[tex]R=\frac{u^2sin2\theta }{\frac{g}{3}}[/tex]
[tex]R=\frac{48.3^2\times \sin52\times 3}{9.8}[/tex]
[tex]R=562.18 m[/tex]
