Question

A stone is thrown horizontally from the top of a 25.00-m cliff. The stone lands at a distance of
40.00 m from the edge of the cliff. What is the initial horizontal velocity of the stone?

Answer 1

The time taken to fall t
s= 1/2 g t^2
25 = 1/2 x 9.8 x t^2
t^2= 5.1
t= 2.26 s

Horizontal velocity = 40/2.26 = 17.6 m/s

Answer 2

The projectiles launch allows to find the answer for the initial velocity of the stone is:

            v = 17.71 m / s

Projectile launching is an application of kinematics for motion in two dimensions, where there is no acceleration on the x axis and the acceleration on the y axis is the gravity acceleration.

In the attached diagram we can see a corner of the movement. Where the x axis is horizontal and the y axis is vertical, where the zero of the system is at the base of the cliff

In this case the stone is thrown horizontally, therefore its initial vertical speed is zero, let's find the time it takes to reach the base of the cliff

           y = y₀ + $v_{oy}$ t - ½ g t²

where y and y₀ are the current and initial position, $v_{oy}$ is the initial vertical velocity, g the acceleration of gravity and t the time

When reaching the bottom its height is zero (y = 0) and the highest part its initial height is y₀ = 25.00 m

           0 =y₀ + 0 - ½ g t²

          t = $\sqrt{\frac{2y_o}{g} }$

Let'se calculate

           $t = \sqrt{ \frac{2 \ 2.25 }{9.8 } }$

           t = 2,259 s

They indicate that the stone fell at a horizontal distance of 40 m,

           x = $v_{ox}$ t

            v_{ox} = $\frac{x}{t}$

           v_{ox}  = $\frac{40}{2.259}$

           v_{ox} = 17.71 m / s

In conclusion, using theprojectiles launch we can find the answer for the initial velocity of the stone is:

            v = 17.71 m / s

Learn more here:   brainly.com/question/12792922