IIT JEE , AIEEE Forum - Physics , Mechanics

piyushsahani

the speed of a projectile at the paximum height is half of its initial speed u. then what is its horizontal range.

astronautameya

at max height, velocity is u cos@

@=60 degree

now, use equation and find
answer is u^2 (root.3) /2g

sinjan.j

At the highest point it is only horizontal velocity,therefore v/2=vcos(theta) i.e. theta is 60 degrees.Now,use the standard formula and put the value of theta obtained.

waterdemon

Here is the solution:

The velocity of the particle at tis max. Hieght is half its

initial velocity.

So At H_max V = U/2

Now when the particle is at maximum height its velocity

in the upward direction is "zero".(The reason why it does

not Go up any further)

So it only has "Horizontal component"

UCos@ = U/2

Cos@ = 1/2

@ = Cos^-1(1/2)

@ = 60⁰.

Now for finding the Range.

R = U²Sin2@/g

R = U²Sin120⁰/g

R = 3(U²/2g)

Hope you find it useful.

plZ rate me for my efforts.

Cheers!!!!!!!!!!!!!

spideyunlimited

velocity at top ( ucos@) = initial velocity (u) / 2

u cos@ = u/2
thus, cos @ = 1/2 and @ = pi/3 . ie.60 degrees

range = u^2.sin2@/g
= u^2. /g X root 3 / 2
= (root 3) u^2 / 2g

putting g as 10 and root 3 as 1.73
= 0.0685 u^2

*PLZ RATE*

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