Question

a cannonball is launched horizontally from the top of a 78.4 - meter high cliff. how much time will it take for the ball to reach the ground? 29/54

Explanation:

Step1: Identify the vertical - motion equation

The vertical - displacement of an object in free - fall is given by the equation $y = y_0+v_{0y}t-\frac{1}{2}gt^2$, where $y$ is the final vertical position, $y_0$ is the initial vertical position, $v_{0y}$ is the initial vertical velocity, $t$ is the time, and $g = 9.8\ m/s^2$ is the acceleration due to gravity. Since the cannonball is launched horizontally, $v_{0y}=0$. The initial height $y_0 = 78.4\ m$ and the final height $y = 0$.

Step2: Substitute values into the equation

The equation becomes $0 = 78.4+0\times t-\frac{1}{2}\times9.8\times t^2$. Simplify it to $4.9t^2=78.4$.

Step3: Solve for $t$

Divide both sides of the equation $4.9t^2 = 78.4$ by $4.9$: $t^2=\frac{78.4}{4.9}=16$. Then take the square - root of both sides. Since time cannot be negative in this context, $t=\sqrt{16}=4\ s$.

Answer:

4 s