Question

Calculate the angle of projection of body if its horizontal range is three times the maximum height. find the time of flight if its velocity of projection is 18m/s​

Answer 1

Answer:

1st method:-

u=18m/s

Let, theta be angle of projection...from horizontal

R=3H

=> (u^2sin 2theta)/g=3(u^2sin^2theta)/2g ( using formulae for range and height)

=> sin 2theta=(3sin^2theta)/2

=>2sin theta cos theta=(3 sin^2 theta)/2

=>tan theta=4/3

So, theta = tan(4/3)^-1 ( tan inverse 4/3) (Ans)

2nd method:-

Direct relation:-

tan theta= 4(max. height)/(Range)

=> tan theta= 4(max. height)/( 3 times max.height)

=> tan theta =4/3

=>theta = tan(4/3)^-1 ( tan inverse 4/3)( Ans)