Question

if a object is thrown vertically up and it passes through two points A andB such that at A its velocity is one third of its initial velocity and at B it has velocity which is one fourth of its initial velocity and the distance between A and B is 28m then find the height of of the maximum projection​

Answer 1

Given:

A object is thrwon vertically up .It passes through 2 points A and B such that at A its velocity is one third its initial velocity and at B it has velocity one forth of initial velocity .

Distance between A and B is 28m.

To Find:

Height of maximum projection

Solution:

Let A is s distance away from ground and B is S distance away from ground.

Use third equation of motion;

$v^{2} - u^{2} = 2as$

distance covered by object upto A = s

initial velocity = u final velocity = v = $\frac{u}{3}$

a = -g        

substitute the values ;

$2(-g)s = (\frac{u}{3} )^{2} - u^{2} \\-2gs = -\frac{8}{9} u^{2}$          (1)

distance covered by object upto B = S

initial velocity = u final velocity = v = $\frac{u}{4}$

$v^{2}- u^{2} = 2as\\v = \frac{u}{4} \\a = -g \\s = S\\\-\frac{15}{16}u^{2} = -2gS$      ( 2)

substract equation 1 by 2 we get;

$2g(S-s)= \frac{15}{16} u^{2} - \frac{8}{9} u^{2} \\S-s = 28 m (given)\\solve;\\u = 106 \frac{m}{s}$

Maximum height of projection:

u = 106 v = 0 a = -g    s = H

$v^{2} - u^{2} = 2as\\0 - (106)(106) = 2(9.8)H\\H = 573.25 m$

Height of maximum projection is 573.25 m.