Science, asked by Princesamrat, 1 year ago

A man is a balloon rising vertically with an acceleration of 4.9 m/s square ,releases a ball 2 second after the balloon had been let go from ground . the greatest height above the ground reached by the ball will be
(a)9.3
(b) 19.6
(c) 24.5
(d)14.7

Answers

Answered by Shubhendu8898
37

Let Ub and Vb be the velocities at t=0 and t=2 for the balloon

And accelaration of balloon(Ab)=4.9=g/2 m/s2

Vb=Ub+(Ab)t       (Ub=0)

from this we get, Vb=g m/s   (Upwards)

and from ,s=(Ub)t+(1/2)(Ab)t^2          (where s is the hieght at which the ball is dropped)

we get s=g meters

when the ball is dropped its initial velocity is gm/s upwards(due to balloon)

hence velocity of ball at ground(using third eq of motion):

V^2=g^2 + 2(g)(g)        (since,,initial vel of ball=g m/s,,,acc due to gravity=g m/s2,,,displacement=g meters)

v^2=3g^2------------------------------------1

If we want to find the greatest hieght(to which it rises)then it should rebound back with same velocity(elastic collision)

therfore ,

v^2=u^2-2(g)(H)---------------------------2   (Where H is the max hieght it reaches)

in this case v=0 since at the greatest hieght the vel of ball is zero

and u^2=3g^2  (from eq--1)

therefore

eq 2 becomes:

3g^2=2gs

ie,s=3/2(g)=14.7m

Answered by pavit15
8

Answer:

(d)14.7

Explanation:

Assuming that the balloon was at rest on the ground

It's initial vertical velocity ( say u ) = 0

The given acceleration of balloon is supposed to be the NET acceleration of the balloon , and that it is obtained after considering the effect of acceleration due to gravity.

So, net acceleration of the balloon ( say a ) = 4.9 m/s² vertically upwards.

Here t = 2s being measured from the instant the balloon starts to rise from the ground.

Using x = u*t + 0.5 * a * t²

Or

x = ( 0*2 +0.5*4.9*2² ) m

= ( 0 + 9.8 ) m = 9.8 m.

When the ball is released from the balloon 2 s after it starts to rise,

the speed of the ball would be same as that of the balloon and in the same direction too.  

Speed of balloon at t = 2s = ( 0 + 4.9*2 ) m/s

=9.8 m/s .

So speed of ball at the point of release = 9.8 m/s vertically upwards ( say v )

So max height reached by the ball,

measured from the point of RELEASE

=v²/2g = { (9.8)²/(2*9.8) } m = 4.9 m

So max. height of ball as measured from the ground = ( 4.9 + 9.8 ) m = 14.7 m .

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