Question

Neglecting the effect of air resistance a stone dropped off a 175-m high building lands on the ground in: A)3s B)4s C)6s D)18s E)36s

Answer 1

Answer:

c) 6s

Explanation: take downwards as positive.

let vf = the velocity of the stone when it hits the ground, vi = 0  be the starting velocity of the stone. r0 = 175m be the height of the building.

(vf)^2 = (vi)^2 + 2g(r-r0)

(vf)^2 = 2g(r0)

     vf =  \sqrt{2g(r0)}  

         = \sqrt{2×9.8×(175)}  

         = 58.6 m/s

then:

vf = vi + g×t

vf = g×t

 t = vf/g

   = (58.6)/(9.8)

   = 6s

Therefore, it will take the stone 6 seconds to reach the gound.

Answer 2

Neglecting the effect of air resistance, time taken for the stone to hit the ground from the top of the high building is 6 seconds.

Hence, Option C) 6s is the correct answer.

Given the data in the question;

Before the stone was dropped, it was initially at rest

• Initial velocity; $u = 0$

• Height of the building; $s = 175m$

Time taken for the stone to hit the ground; $t = \ ?$

Using using the Second Equation of Motion:

$s = ut + \frac{1}{2}at^2$

Where s is the height of the building, u is the initial velocity, t is the time and a is acceleration due to gravity (Since the stone is under gravity; $a = g = 9.8m/s^2$)

We substitute our values into the equation

$175m = [0 * t ] + [ \frac{1}{2}\ *\ 9.8m/s^2\ *\ t^2} ]\\\\175m = \frac{1}{2}\ *\ 9.8m/s^2\ *\ t^2} \\\\175m = 4.9m/s^2\ *\ t^2\\\\t^2 = \frac{175m}{4.9m/s^2} \\\\t^2 = 35.714s^2\\\\t = \sqrt{35.714s^2} \\\\t = 5.97s\\\\t = 6s$

Neglecting the effect of air resistance, time taken for the stone to hit the ground from the top of the high building is 6 seconds.

Hence, Option C) 6s is the correct answer.

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