Question

A particle is projected from ground in vertical direction at t = 0. At t = 0.8s, it reaches h = 14 m. It will again come to same height at t =

Answer 1

Given:

Particle projected from ground in vertical direction at t = 0. At t = 0.8 sec , it reaches 14 m.

To find:

Time taken by the particle to again come to the same height.

Calculation:

Let initial velocity be u :

s = ut + ½ at²

=> 14 = u ×(0.8) + ½× (-10) ×(0.8)²

=> 0.8 u = 14 + 3.2 = 17.2

=> u = 21.5 m/s

So, the time taken to reach max height be T

v = u + at

=> 0 = 21.5 + (-10)T

=> T = 2.15 sec.

Let max height reached be H:

v² = u² + 2as

=> 0 = (21.5)² + 2 × (-10) × H

=> H = 462.25/20

=> H = 23.11 m.

So, while descending , the object will travel

(23.11 - 14) = 9.11 m to reach the same initial height.

Time taken to travel 9.11m while descending

9.11 = ½ gt²

=> t² = (2 × 9.11)/10

=> t² = 1.822

=> t = 1.3 sec.

So total time required to again reach that point will be :

T net = 2.15 + 1.3 = 3.45 sec.

Final answer :

$\boxed{ \bold{T \: net = 3.45 \: sec}}$

Answer 2

first, make a quadratic equation

using,

s=ut + 1/2at ^2

you will get 2 roots, I.e.2 values for t

t1 and t2

now you know t1=0.8

and t2 ×t1=c/a

(product of roots is c/a)

so put values and t2= 7/2

hope this helps