Question

A ground to ground projectile has a range of 200 m and time of flight of 10s and ratio of its maximum to minimum velocity during its flight is r then r is b/w

(a) 1.5-2 (b) 2.5 - 3

Answer 1

b

Answer 2

@ajprincess

Answer 3

@maheshmeghwal9 @Ishaan94

Answer 4

@phi

Answer 5

ratio of max velocity with wt?

Answer 6

max-min

Answer 7

oh !

Answer 8

\[\frac{u^2 \sin 2 \theta }{g}=200. \]\[\frac{2u \sin \theta }{g}=10.\]

Get u & theta.

then

Maximum velocity is "u".
Minimum velocity is "u cos theta". {at the highest point.}

So ratio will be \[\frac{u}{u \cos \theta}.\]\[\implies \frac{1}{\cos \theta}.\]

Answer 9

I think it must be like this but nt sure:)
but u can try my way:)

Answer 10

wt answer do u get after solving ?

Answer 11

I got r= 0.05...just divide eqn 1 by eqn 2 in maheshmeghwal9's post...I don't know why the answer isn't in either one of the opns..

Answer 12

The answer should be (b) 2.5 - 3

Answer 13

Main Problem is that
wt is max velocity
& min. velocity
can anyone help us Plz?????

Answer 14

I m nt sure with my values of max velocity & min. velocity :/

Answer 15

You're right about max and min velocity... As if you resolve into the components, then horiz comp remains the same alsways.., and vertical comp of velocity keeps decreasing uptil max height is reached..so till then, the velocity will decrease and after that, because the parabola is symmetric, the velocity will increase downwards and end up with the same initial value( talking about magnitudes here)..

Answer 16

thanx @aish_premrenu :D
but why we aren't getting answer:/

Answer 17

can anyone help us ?

Answer 18

man ... i tried of your projectiles. find the vertical component of velocity ...

Answer 19

i think
that is tired*