Question

a shell is fired at an angle 40° at the horizontal with a velocity of 300m/s. Find the maximum height and the horizontal range attained​

Answer 1

Answer:

So this is an example of oblique projectile thrown from the ground.

Now, as per the question,

The initial velocity is 300 m/s.

And angle of throw is 40°

therefore u= 300 m/s

and theta is 40 °

You have to remember the formula of maximum height and range in a projectile.

The max height attained is given by

H = [u^2 sin( theta)]/ 2g. (g - gravity)

and the range is obtained by

R = [u^2 sin ( 2 theta)]/g.

Now look at the attached photo,

please put the value of sin (80°) and

sin^2(40°).

hope you will be able to understand.