Answer:81.24 m/s
Explanation:
Given
Horizontal displacement([tex]R_x[/tex])=500
Angle of projection[tex]=24 ^{\circ}[/tex]
Let u be the launching velocity
and horizontal range is given by
[tex]R_x=\frac{u^2sin2\theta }{g}[/tex]
[tex]500=\frac{u^2sin48}{9.81}[/tex]
[tex]u^2=\frac{500\times 9.81}{0.7431}[/tex]
[tex]u^2=6600.32854[/tex]
[tex]u=\sqrt{6600.32854}=81.24 m/s[/tex]
