A stone is thrown upwards with a speed u from the top of a tower.It reaches the ground with a velocity 2u.The height of tower
Answers
Answered by
0
Intially stone has potential energy only mhu later potential energy convert into kinetic energy 1/2m(2u)(2u) equate both and h=2u
Answered by
1
first of all we have to calculate the hight when the stone reach at maximum
so here u = u a = -g and at maximum hight it's velocity become zero
so V² - u² = 2as
0 - u² = 2(-g)H
H = u²/2g...............1
now case 2 when stone come in upwards direction
in this case u = 0 v = 2u and s = h
v² - u² = 2as
(2u)²- 0 = 2(-g)h
4u² = - 2gh
h = -2u²/g...............2
but we need hight of tower so eq. 2-1
hight of tower = 2u²/g - u²/2g
= 3u²/g
so here u = u a = -g and at maximum hight it's velocity become zero
so V² - u² = 2as
0 - u² = 2(-g)H
H = u²/2g...............1
now case 2 when stone come in upwards direction
in this case u = 0 v = 2u and s = h
v² - u² = 2as
(2u)²- 0 = 2(-g)h
4u² = - 2gh
h = -2u²/g...............2
but we need hight of tower so eq. 2-1
hight of tower = 2u²/g - u²/2g
= 3u²/g
Similar questions