Question

A body is dropped from rest at a height of 80m how long does it takes to reach the ground ?(g=10ms square ignore air resistance)

Answer 1

Explanation:

given body is dropped from rest so initial velocity(u) is

zero

u =0 , height = 80m and time taken to reach ground = ?

g = 10m/s^2

here acceleration is nothing but gravity because when it is dropped from a certain height it accelerates du to gravity so in this case acceleration is 10m/s^2

we have u , h , g and t can be found by using equations of motion

using 3rd equation of motion,

s = ut + at^2/2

80 = 0 (t) + 10(t^2)/2

80 = t + 10t^2 /2

80 = 2t + 10t^2/2

160 = 10t^2 + 2t

0 = 10t^2 + 2t - 160

using the formula for solving a quadratic equation

a = 10 , b= 2 , c = -160

- b +root overb^2 - 4ac / 2a (or ) -b - root over b^2-4ac/2a

= - 2 + root over 4 + 4(10)(160) / 2(10)

(or) -2-root over 4 + 4(10)(160) / 2(10)

= -2 + root over 6400 / 20

(or) -2 - root over 6400 / 20

= -2 + 80 / 20 (or) -2-80 / 20

= 78/20 ( or ) -82/ 20

= 3.9 (or) - 4.1

time can't be negative so 3.9 is the time required by the body to reach the ground