Answer:
The answer is "8.28 m".
Explanation:
[tex]\to u \cos 53^{\circ}=6 \ \text{therefore the horizontal velocity is constant alaways}\\\\\therefore[/tex]
[tex]u= \frac{6 \times 5}{3}= \frac{30}{3}= 10\ \frac{m}{s}\\\\[/tex]
In the projectfile when vertical displacement is 1 then 5 m comes at farther sides
Using
[tex]S=ut-\frac{1}{2} gt^2\\\\1=u\sin 53^{\circ}\times t-5t^2\\\\1=8t-5t^2\\\\5t^2-8t+1=0\\\\ t=\frac{8\pm \sqrt{64-20}}{10} \\\\=\frac{8\pm 2\sqrt{11}}{10} \\\\t_2=\frac{8+2\sqrt{11}}{10} \\\\[/tex]
[tex]Distance=u \cos 53^{\circ}\times t_2\\\\[/tex]
[tex]=6 \times \frac{8\pm 2\sqrt{11}}{10} \\\\=\frac{24\pm 6\sqrt{11}}{5} \\\\= 8.28 \ m[/tex]
