Question

A shell is fired at an angle 30 degree to the horizontal with a velocity 392 m/s. Find the total time of flight horizontal range and maximum height attained?

Answer 1

If a shell is projected with an initial velocity of `u` at an angle `theta` w.r.t to the horizontal,

It's range is given by the formula `(u^2 sin 2theta)/g`

Total time of flight as `(2u sin theta)/g`

And,maximum height as `(u sin theta)^2/(2g)`

Now,given, `u=392 m/s` and `theta=30`

So,putting the values we get,

Total time of flight = `39.96 s`

Range = `13565 m`

Maximum height reached = `1958 m`

If you need the derivation of the formulae , let me know

Answer 2

See the explanation below

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=392sin(1/6pi)ms^-1`

Applying the equation of motion

`v^2=u^2+2as`

At the greatest height, `v=0ms^-1`

The acceleration due to gravity is `a=-g=-9.8ms^-2`

Therefore,

The greatest height is `h_y=s=(0-(392sin(1/6pi))^2)/(-2g)`

`h_y=(392sin(1/6pi))^2/(2g)=1960m`

The time to reach the greatest height is `=ts`

Applying the equation of motion

`v=u+at=u- g t`

The time is `t=(v-u)/(-g)=(0-392sin(1/6pi))/(-9.8)=20s`

The total time of flight is `=40s`

Resolving in the horizontal direction `rarr^+`

The velocity is constant and `u_x=392cos(1/6pi)`

The range in the horizontal direction is

`s_x=u_x*t=392cos(1/6pi)*2*20=13579.3m`

graph{-(19.6/(3*392^2))x^2+(1/sqrt3)x [-293, 14340, -2480, 4844]}