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Grade: 11

                        

A particle is projected upwards . the times corresponding to height h while ascending and while descending are 4s and 6s respectively . Find the velocity of projection

3 years ago

Answers : (1)

Arun
24737 Points
							
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
hence velocity of projection in question v = g(4+6)/2
v = 5g
3 years ago
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