Prove that there are two values of time for which a projectile is at the same height. Also show that the sum of these two times is equal to the time of flight.
Answers
Answered by
68
In a projectile along the y-axis the motion is an accelerated, equation of motion in Y direction is:
Y = (usinA) t - 1/2gt^2
Rearrange the terms to get
t^2 - 2tusinA/g + 2y/g = 0
Which is a quadratic in t
Hence we will have two solutions for t for a given y(height)
The sum of the roots will be
2usinA/g which is equal to time of flight
Hope it helps
Y = (usinA) t - 1/2gt^2
Rearrange the terms to get
t^2 - 2tusinA/g + 2y/g = 0
Which is a quadratic in t
Hence we will have two solutions for t for a given y(height)
The sum of the roots will be
2usinA/g which is equal to time of flight
Hope it helps
Answered by
20
Answer:
initially v=vcos\theta i+vsin\theta j
so to be at height h
we set up this eq
h=vsin\thetat-1/2at^2
whose sum of roots is -b/c=2vsin\theta/g
which is the time of flight
so if when going up height h time is t
then on coming down t’=Time of flight -t
Explanation:
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