Answer:
[tex]R_{max} = 51.84 m[/tex]
Explanation:
When we shoot the dart upwards the time taken by the dart to go straight up and again come back is given as
[tex]t = 4.6 s[/tex]
here we can say
[tex]\Delta y = 0 = v_i t + \frac{1}{2}at^2[/tex]
[tex]0 = v_i t - \frac{1}{2}(9.8)t^2[/tex]
put t = 4.6 s then we have
[tex]v_i = 22.54 s[/tex]
Now in order to find the maximum range we can say
[tex]R = \frac{v^2 sin2\theta}{g}[/tex]
so in order to have maximum range we can say
[tex]\theta = 45 degree[/tex]
[tex]R_{max} = \frac{v^2}{g}[/tex]
[tex]R_{max} = \frac{22.54^2}{9.8}[/tex]
[tex]R_{max} = 51.84 m[/tex]