Question

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How

many seconds will have elapsed before the cannon ball strikes the ground?

Answer 1

It takes approximately 3.55 seconds for a cannonball fired horizontally at 243 m/s from a 62 meter tall cliff to hit the ground, considering only its vertical motion affected by gravity.

Step-by-step explanation:

The student's question involves calculating the time it takes for a cannonball to hit the ground when fired horizontally from a cliff. This is a classic example of projectile motion in physics, where the vertical and horizontal motions can be treated independently.

To find the time taken for the cannonball to reach the ground, we only need to consider the vertical motion. Since it's fired horizontally, the initial vertical velocity is 0 m/s. We use the formula for the motion under constant acceleration due to gravity (g = 9.81 m/s²), which is h = 0.5 * g * t², where h is the height of the cliff and t is time in seconds. Rearranging for t gives t = √((2 * h) / g).

Substituting the height of the cliff (h = 62 m) into the equation results in t = √((2 * 62 m) / 9.81 m/s²) . Calculating this we get the time t ≈ 3.55 s.

Therefore, it will take approximately 3.55 seconds for the cannonball to strike the ground after being fired.

Answer 2

Horizontal component of ball, u = 0 m/s.

Height from which it thrown, h = 62 m.

By equation of motion :

`s=ut+(at^2)/(2)`

Putting all given equation, we get :

`62=0* t+(10t^2)/(2)\\\\t=√(12.4)\ s\\\\t = 3.52\ s`

Therefore, time taken is 3.52 s.

Hence, this is the required solution.