Two particles projected form the same point with same speed u at angles of projection `alpha and beta` strike the horizontal ground at the same point. If `h_1 and h_2` are the maximum heights attained by the projectile, R is the range for both and `t_1 and t_2` are their times of flights, respectively, then

Options

A. `alpha+beta=90^(@)`
B. `t_(1)=t_(2) tan alpha`
C. `R=4 sqrt(h_(1)h_(2))`
D. `(h_(1))/(h_(2))tan^(2) alpha`

Answer

A, B, C, D

Explanation

(a,b,c,d): When particle are projected with the same speed at angles `theta and (90^(@)-theta)`, then range of projectile is found to be the same . Hence `(alpha +beta)=90^(@)` and option (a) is correct .
`(2u sin alpha)/((t_(1))/(t_(2))=((g)/(2u sin beta))/(g))=(sin alpha)/(sin beta)`
`rArr (t_(1))/(t_(2))=(sin alpha)/(sin (90 -alpha))=(sin alpha)/(cos alpha)=tan alpha`
`rArr t_(1)=t_(2) tan alpha`, hence option (b) is also correct .
Also , `R=(u^(2) sin 2alpha)/(g)`
`rArr h_(1)=(u^(2) sin^(2)alpha)/(2g)`
`rArr h_(2)=(u^(2)sin^(2)beta)/(2g)=(u^(2)sin^(2)(90^(@)-alpha))/(2g)=(u^(2) cos^(2) alpha)/(2g)`
We can now write the following ,
`sqrt(h_(1)h_(2))=(u^(2)sin alpha cos alpha)/(2g)=(u^(2) (2 sin alpha cos alpha)/(4g))`
`rArr(u^(2) (2sin alpha cos alpha))/(g)=4sqrt(h_(1)h_(2))`
`rArr (u^(2) sin 2 alpha)/(g)=4sqrt(h_(1)h_(2))`
`rArr R=4sqrt(h_(1)h_(2))`
Hence option (c) is also correct. Further we can see that `(h_(1))/(h_(2))=tan^(2)alpha`.
Hence option (d) is also correct.