A ball is dropped from a height h and covers h/2 distance in the last second, calculate for how long it remains in air
Answers
Answered by
0
The total distance covered in time t by the body to reach the ground is given by,
h1=1/2gt^2
=1/2×10×t^2
=5t^2
The body moves for (t−1) second before the beginning of the last second.
The velocity of the body at an instant (t−1) second isv=g(t−1)=10(t−1)
This is the initial velocity for the motion in the last one second.
The distance covered in this one second is
h^2=10(t−1)×1+1/2×10×1^2
=10(t−1)+5=10t−5
It is given that,
h^2=h1/2
Therefore,10t−5=5t^2/2
or
t^2−4t+2=0
or
t=(2±2)−−√ s
h1=1/2gt^2
=1/2×10×t^2
=5t^2
The body moves for (t−1) second before the beginning of the last second.
The velocity of the body at an instant (t−1) second isv=g(t−1)=10(t−1)
This is the initial velocity for the motion in the last one second.
The distance covered in this one second is
h^2=10(t−1)×1+1/2×10×1^2
=10(t−1)+5=10t−5
It is given that,
h^2=h1/2
Therefore,10t−5=5t^2/2
or
t^2−4t+2=0
or
t=(2±2)−−√ s
Similar questions