Find angle of projection with the horizontal in terms of maximum height attained and horizontal range.

Question

Find angle of projection with the horizontal in terms of maximum height attained and horizontal range. (A)  tan -1 (2 H/R) (B)  tan -1 (4 R/H) (C)  tan -1 (4 H/R) (D)  tan -1 (H/R)

Tardigrade Admin · Accepted Answer

Maximum height, H=(u2 sin 2 θ/2 g) .....(i) Horizontal range, R=(u2 sin 2 θ/g) .....(ii) Dividing Eq. (i) by Eq. (ii), we get (H/R)=( tan θ/4) θ= tan -1 (4 H/R)

Q. Find angle of projection with the horizontal in terms of maximum height attained and horizontal range.

$tan^{- 1} \frac{2 H}{R}$

$tan^{- 1} \frac{4 R}{H}$

$tan^{- 1} \frac{4 H}{R}$

$tan^{- 1} \frac{H}{R}$

Maximum height, $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ .....(i)
Horizontal range, $R = \frac{u ^{2} s i n 2 θ}{g}$ .....(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{H}{R} = \frac{t a n θ}{4}$
$θ = tan^{- 1} \frac{4 H}{R}$

Q. Find angle of projection with the horizontal in terms of maximum height attained and horizontal range.

tan−1R2H​

tan−1H4R​

tan−1R4H​

tan−1RH​

Maximum height, H=2gu2sin2θ​ .....(i) Horizontal range, R=gu2sin2θ​ .....(ii) Dividing Eq. (i) by Eq. (ii), we get RH​=4tanθ​ θ=tan−1R4H​

$tan^{- 1} \frac{2 H}{R}$

$tan^{- 1} \frac{4 R}{H}$

$tan^{- 1} \frac{4 H}{R}$

$tan^{- 1} \frac{H}{R}$

Maximum height, $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ .....(i)
Horizontal range, $R = \frac{u ^{2} s i n 2 θ}{g}$ .....(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{H}{R} = \frac{t a n θ}{4}$
$θ = tan^{- 1} \frac{4 H}{R}$