Question

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80-m high cliff. The shell strikes the ground 1330 m from the base of the cliff. The drawing is not to scale. What is the magnitude of the velocity of the shell as it hits the ground?

Answer 1

`V = 331.6946\ m/s`

Step-by-step explanation:

First let's find the time that takes the shell to hit the ground (height zero).

To find this time, we can use the equation:

`S = So + Vo*t + at^2/2`

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time. Then, for the vertical movement of the shell, we have that:

`0 = 80 + 0*t - 9.81*t^2/2`

`9.81*t^2/2 = 80`

`t^2 = 160/9.81 = 16.31`

`t =4.0386\ seconds`

Now, to find the horizontal speed, we use the equation:

`S = So + V*t`

Then, for the horizontal movement, we have:

`1330 = 0 + V_h * 4.0386`

`V_h = 1330/4.0386 =329.32\ m/s`

Now we need to find the vertical speed, using:

`V = Vo + a*t`

`V_v = 0 - 9.81*4.0386`

`V_v = -39.6187\ m/s`

Finally, to find the magnitude of the velocity, we have:

`V = √(V_h^2 + V_v^2)`

`V = √(329.32^2 + (-39.6187)^2)`

`V = 331.6946\ m/s`