Question

A rock is thrown off a cliff with a speed of 0.7 m/s downward. How far will it fall after 3 seconds have elapsed?

Who ever answer it right will be marked as Brainlist answer

Answer 1

Time it will fall = 3 s

Initial velocity = 0.7 m/s

Acceleration due to gravity = + 10 m/s² [Down]

We know,

h = ut + ½gt²

⟹ h = (0.7 m/s)(3 s) + ½(10 m/s²)(3 s)²

⟹ h = 2.1 m + (5 m/s²)(9 s²)

⟹ h = 2.1 m + 45 m

⟹ h = 47.1 m

∴ It will cover a distance of 47.1 m before collision with the surface.

@Psyclone

Answer 2

Given: A rock is thrown off a cliff with a speed of 0.7 m/s downward.

To find: Distance it has fallen after 3 seconds has elapsed.

Solution:

• Speed of an object is the distance travelled by it in unit time.
• It is given by the formula,

        $s = \frac{d}{t}$

• Here, s is the speed of the rock at which it is thrown, d is the distance travelled by it and t is the time period.
• On replacing the terms with the values given in the question, the equation obtained is,

        $0.7ms^{-1} = \frac{d}{3s}$

• On rearranging, the value of d is found to be,

        $d = 2.1m$