Question

the maximum height attained by a shell is 80 m & the horizontal range is 320m what is the velocity & angle of projection

Answer 1

Answer:

The velocity is 40.2 m/s and angle of projection is 5.71°

Explanation:

According to the problem the height,h the shell has attained is 80 m and the

horizontal range is 320 m

Let θ is angle of projection

h = v^2sin^2θ/2g=0.80 [ where v is the initial velocity]

Assuming g = 10 m/s^2

v^2sin^2θ=0.80 ×2×10

=> v^2sin^2θ = 16

=> v^2 = 16/sin^2θ

Now,  k = v^2sin^2θ/g=320

Now putting all the values,

               16/sin^2θ x (2sinθcosθ)/ g = 320

(16/sinθ)(2cosθ)=320×g

32cot θ =3200

cot θ=10

1/tan θ=10

tan θ=0.1

θ=tan−1(0.1)= 5.71°

Therefore,

v^2=16/sin^2( 5.71°)

v=4sin( 5.71°)

  =40.2 m/s