A body is projected down with velocity 5m/S from a height of 60 m find the time of descent
Answers
Concept:
The third kinematic equation is illustrated as, h = ut + 1/2 gt²
Given:
The velocity of the body when downward projected = 5m/s
Height of the body = 60 m
Find:
We need to determine the time of descent.
Solution:
We know the third kinematic equation in terms of acceleration due to gravity as- h = ut + 1/2 gt² where u is the initial velocity, t is the time and g is the acceleration due to gravity.
We have height as h = 60 m
Taking g as 10m/s²
The third kinematic equation h = ut + 1/2 gt² becomes-
60 = 5t + 1/2(10)t²
60 = 5t + 5t²
0 = 5t + 5t² - 60
We can solve the above equation using the quadratic equation formula which becomes an as-
(-b ± √b² - 4ac)/2a
It becomes, (- 5 ± √5²- 4×5×-60) /2×5
It becomes, (- 5 ± √25+1200)/10
Thus we get two terms for t
t = -5 + 35 / 10 and t = - 5 - 35 /10
t = 3 seconds and t = - 4 sec
As the time cannot be negative so the time becomes 3 seconds.
Thus, the time of descent is 3 seconds.
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