Question

A cannon standing on the top of a cliff, 40m high fires a cannon ball horizontally out to sea with a muzzle speed of 140m/sec. How far out to sea does the ball go ?

Answer 1

Answer:$280\sqrt{2}m$

Explanation:

We have,

Speed of muzzle in horizontal direction, u = 140 m/s .

Height of cliff, H= 40 m

Time taken to fall onto sea, t = ?

Initial Velocity in vertical downward direction, u_y = 0

Then,

Using Newton's Equation of motion,

$S_y = u_yt +\frac{1}{2}a_yt^2 \\ \\=>-H=0*t-\frac{1}{2}*g*t^2\\ \\=>-40=-0.5*10*t^2\\ \\=>t=2\sqrt{2}s$

Now, we will find distance covered in t secs .

That is,

$D=u_xt\\ \\=> D=140*2\sqrt{2}=280\sqrt{2}m$

Therefore,

Distance covered by ball, d = 280√2 m.

Answer 2

Answer:

280*2^(-2)  

Explanation:

We know that in horizontal projection the formula of horizontal range is u*(2h/g)^-2

Here u=140 m/s    and     h = 40 m    

So if we take g = 10 m/s^2

Horizontal Range = 140*(40/10)^-2

                               =140*4^(-2)

                               = 280*2^(-2)  

Hope it helps.