A particle is projected upwards. The times corresponding to height h while ascending and descending are t1 and t2 respectively. Find the velocity of projection?
Given: A particle is projected upwards. Height = h time of ascent = t1 time of descent =t2. a=-g let u be the initial velocity. u=? By using second equations of motions, S=ut+1/2 at² h=ut-1/2gt² 2h=2ut-gt² gt²-2ut-+2h=0 This is quadratic equation which corresponds in t which shows that for a given h there are two values of t [t1 and t2] By comparing the above equation with quadratic equation, ax²-bx+c=0 we have : sum of roots =t1+t2= -b/a= -(-2u)/g=2u/g Product of roots=t1t2=c/a=2h/g ∴ sum of roots, t1+t2=2u/g ∴u=g(t1+t2)/2 ∴Velocity of projection is g(t1+t2)/2
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Given:A particle is projected upwards.
Height = h
time of ascent = t1
time of descent =t2.
a=-g
let u be the initial velocity.
u=?
By using second equations of motions,
S=ut+1/2 at²
h=ut-1/2gt²
2h=2ut-gt²
gt²-2ut-+2h=0
This is quadratic equation which corresponds in t which shows that for a given h there are two values of t [t1 and t2]
By comparing the above equation with quadratic equation,
ax²-bx+c=0
we have :
sum of roots =t1+t2= -b/a= -(-2u)/g=2u/g
Product of roots=t1t2=c/a=2h/g
∴ sum of roots, t1+t2=2u/g
∴u=g(t1+t2)/2
∴Velocity of projection is g(t1+t2)/2