Recent content by AlphaA

Thank you so much .. Sir... I got it !... It was really simple and I was stuck at it like anything... Thank you so much for helping me out.

Oh...THANKYOU soooooooooooooooo much... I got it! I can now see it clearly...yaa...I've undersood it now... Once again...THANKS A LOT ! It was really very sweet of you to guide me ... Gratitude.

My doubt is the underlined portion only. If the rocket moves against gravity then why should acceleration due to gravity be added to the initial acceleration?

The example goes on like this : A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m/s^2 . Find the initial thrust(force) of the blast. Take g=9.8 m/s^2 Solution (as given in my book) : The rocket moves up against gravity with an acceleration of 5.0...

But as the bookish explanation says, we add the positive value of acceleration due to gravity and not the negative one... So here adding it means adding it only... That's why I am confused... Need some help and guidance

If an object say a rocket is thrown up in the sky with an additional acceleration say ' x ' , then why do we add the value of acceleration due to gravity i.e 9.8 m/s^2 to the acceleration ' x' in order to find the total acceleration... Since vector of acceleration due to gravity is directed...

Sir, my problem is this only. I get horribly confused while taking a sign convention for equations of motion in numericals of projectile motion. Suppose if i choose a vector going upward to be positive and vector going downward to be negative. Then in this situation,if a package is dropped from...

1. v=u+at 2. s=ut+0.5at( t raised to the power 2) 3. v(raised to the power 2)=u(raised to the power 2)+ 2as where v=final velocity u=initial velocity a= acceleration t= time s= distance/displacement/height

< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >[/color] A balloon going upwards with a velocity of 12 m/s, is at a height of 65m from the Earth at any instant. Exactly at this instant a packet drops from it. How much time will the packet take...