Question

Calculate the least velocity of projection required to give

Answer 1

Answer:81.24 m/s

Explanation:

Given

Horizontal displacement($R_x$)=500

Angle of projection$=24 ^{\circ}$

Let u be the launching velocity

and horizontal range is given by

$R_x=\frac{u^2sin2\theta }{g}$

$500=\frac{u^2sin48}{9.81}$

$u^2=\frac{500\times 9.81}{0.7431}$

$u^2=6600.32854$

$u=\sqrt{6600.32854}=81.24 m/s$