Question

A bullet is fired from horizontal ground at some angle passes through the point(3R/4,R/4),where R is the range of the bullet. assume point of the fire to be origin and the bullet moves in x-y plane with x-axis horizontal and y-axis vertically upwards. angle of projection is (alpha *pie)/180 radian. find alpha.​

Answer 1

Answer:

Explanation:

As the given points are given in the x-y plane hence thepoints will also be in x,y. Thus, let x=3/4R and y =R/4.

Taking x= 3R/4 , 3R/4 = ucosθ*t.

So, t=3R / 4ucosθ.

Now, by using equation of laws of motion,

y=usinθ*t - 1/2g$t^{2}$

R/4 = usinθ * 3R / 4ucosθ  -  1/2g* (3R / 4ucosθ)^2.

Putting the value of  R= u^2*sin2θ/g and sin2θ=2sinθcosθ

Which , on solving we get 3sin2θ=cos^2θ.

Which, we get as 6sinθ=cosθ.

θ= $tan^{-1}$(0.16).

(alpha *pie)/180 = 9.09.

alpha = 180*9.09/pie.