Question

A bullet fired from a rifle attains maximum height of 25 m and crosses a range of 200m. Find the angle of projection.



Plz friends give me correct answer​

Answer 1

Given :-

$H = 25 m$

$R = 200 m$

$g=9.8 \frac{m}{s^2}$

Required to find ;-

θ = ?

Procedure ;-

Let us recall the maximum height achieved by projectile as follows :

$H=\frac{u^2sin^2 \theta}{2g}$

Let us plug in all that we know.

$25=\frac{u^2sin^2 \theta}{2*9.8}$

$490=u^2sin^2 \theta$ --------- 1

Now let us recall the range achieved by projectile as follows :

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

Let us now plug in all that we know.

$200=\frac{u^2sin2\theta}{9.8}$

$1960=u^2sin2\theta$ ---------- 2

Now let us divide 2 by 1 .

$\frac{1960}{490} = \frac{sin^2\theta}{sin2\theta}$

$4 = \frac{sin\theta}{2cos\theta}$

$8=tan\theta$

$\theta=tan^{-1}(x)$

∴Angle of projection is  $tan^{-1}(x)$ .

Answer 2

Given :-

H = 25 m

R = 200 m

Required to find ;-

θ = ?

Procedure ;-

Let us recall the maximum height achieved by projectile as follows :

Let us plug in all that we know.

25={u^2sin^2 \theta}{2*9.8}25=2∗9.8u2sin2θ

490=u^2sin^2 \theta490=u2sin2θ --------- 1

Now let us recall the range achieved by projectile as follows :

R={u^2sin2\theta}{g}R=gu2sin2θ

Let us now plug in all that we know.

200={u^2sin2\theta}{9.8}200=9.8u2sin2θ

1960=u^2sin2\theta1960=u2sin2θ ---------- 2

Now let us divide 2 by {1960}{490} = \frac{sin^2\theta}{sin2\theta}4901960=sin2θsin2θ

4 = \frac{sin\theta}{2cos\theta}4=2cosθsinθ

8=tan\theta8=tanθ

\theta=tan^{-1}(x)θ=tan−1(x)

∴Angle of projection is  tan^{-1}(x)tan−1(x)

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