An object falls a distance D from rest. The object travels a distance 0.51 D in last 2 seconds. The time taken by the object to cover the distance D is
Answers
Given : An object falls a distance D from rest. The object travels a distance 0.51 D in last 2 seconds.
To find : The time taken by the object to cover the distance D
Solution : initial velocity, u = 0
let v is the velocity of object after falling 0.49D distance.
using formula, v² = u² + 2as
⇒v² = 0² + 2(-9.8 m/s²)(-0.49D)
⇒v² = 9.8 × 0.98D
⇒v = 0.98 √(10D)
time taken to travel last 0.51D distance is 2sec.
here initial velocity, u = v = 0.98√(10D) , t = 2 sec
using formula, s = ut + 1/2 at²
⇒0.51D = 0.98√(10D) × 2 + 1/2 × 9.8 × 2²
⇒0.51D = 1.96√(10D) + 19.6
⇒0.51D = 6.2√D + 19.6
⇒0.51√D² - 6.2√D - 19.6 = 0
let √D = s
⇒0.51s² - 6.2s - 19.6 = 0
⇒s = 14.76
⇒√D = 14.76
⇒D = (14.76)²
now time taken to cover the distance D , t = √(2D/g)
= √{2 × (14.76)²/9.8}
= 14.76√{1/4.9}
= 6.667 sec
Therefore time taken by the object to cover the distance D is 6.667 sec .
Answer is 30/7 s that is 4.2 s we meed to always choose answers approximately.
Explanation:
check out the pic.. hope it helps you.
