Question

Calculate the least velocity of projection required to give

amissile a horizontal displacement of 500m if the angle
ofprojection is 24 degrees?

Answer 1

Answer:81.24 m/s

Step-by-step 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=(u^2sin2\theta )/(g)`

`500=(u^2sin48)/(9.81)`

`u^2=(500* 9.81)/(0.7431)`

`u^2=6600.32854`

`u=√(6600.32854)=81.24 m/s`