Answer:
Explanation:
Maximum height of a projectile
[tex]H_{max}=\frac{u^2\sin ^2\theta }{2g}[/tex]
(1)[tex]u=30.7\ m/s[/tex]
[tex]\theta =35.7^{\circ}[/tex]
[tex]H_1=\frac{30.7^2\times \sin ^2(35.7)}{2\times 9.8}[/tex]
[tex]H_1=16.37\ m[/tex]
(2)[tex]u=47.9\ m/s[/tex]
[tex]\theta =33.2^{\circ}[/tex]
[tex]H_2=\frac{47.9^2\times \sin ^2(33.2)}{2\times 9.8}[/tex]
[tex]H_2=35.09\ m[/tex]
(3)[tex]u=21.8\ m/s[/tex]
[tex]\theta =67.7^{\circ}[/tex]
[tex]H_3=\frac{21.8^2\times \sin ^2(67.7)}{2\times 9.8}[/tex]
[tex]H_3=20.75\ m[/tex]
(4)[tex]u=7.7\ m/s[/tex]
[tex]\theta =62.1^{\circ}[/tex]
[tex]H_4=\frac{7.7^2\times \sin ^2(62.1)}{2\times 9.8}[/tex]
[tex]H_4=2.36\ m[/tex]
Therefore height reached by pumpkin can be arranged as
[tex]H_4<H_1<H_3<H_2[/tex]