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

Answer:

$\underline{\purple{\ddot{\Mathsdude}}}$

Answered by Rohith kumar maths dude :-

Given:-

• H=25m

• R=200m

To prove:-

• Angle of projection

Proof;-

• H=25=u^2sin^2theta/2g ------(i)

• R=200m= u^2sin 2 theta/g -------(ii)

●Now dividing 1 and 2 equation,

♧♧we get,

u^2sin^2theta/2g × g/u^2sin2 theta=25/200

sin^2theta/2 ×1/2sin theta=1/8

sin^2theta/sin 2 theta =1/4

sin theta/costheta =1/2

♧♧Tan theta=1/2.

●we write it as,

♧♧tan theta= 26.6 degree

Hope it helps u mate.

♧♧Thank you.

Answer 2

Explanation:

Given:-

H=25m

R=200m

To prove:-

Angle of projection

Proof;-

H=25=u^2sin^2theta/2g ------(i)

R=200m= u^2sin 2 theta/g -------(ii)

Now dividing 1 and 2 equation,

we get,

u^2sin^2theta/2g × g/u^2sin2 theta=25/200

sin^2theta/2 ×1/2sin theta=1/8

sin^2theta/sin 2 theta =1/4

sin theta/costheta =1/2

♧♧Tan theta=1/2.

we write it as,

tan theta= 26.6 degree

Hope it helps u you

♧♧Thank you.

From kanav