Question

jee mains related questions​

Answer 1

Question: A ball is released from the top of the tower of height h. It takes time T to reach on the ground. Height of the ball in time T/2 is nh/4. Find n.

Answer:

3

Explanation:

As the body is 'released', its initial velocity(u) is 0.

Height of the tower(S)= h

Time(t) = T ; a = g

Using the equations of motion:

=> S = ut + 1/2 at²

=> h = 0 + 1/2 gT²

=> h = 1/2 gT² ...(1)

In t = T/2 (from the top of the tower)

h' = (0)(T/2) + 1/2 g(T/2)²

h' = gT²/8 ...(2)

Divide (2) by (1), we get:

=> h'/h = (gT²/8) / (gT²/2)

=> h'/h = 1/4

=> h' = h/4

Therefore, height from ground is:

Height from ground = h - h/4 = 3h/4

Compare this with nh/4,

3h/4 = nh/4

3 = n

Answer 2

Question:-

A ball is released from the top of a height h. It takes time T to reach on the ground. Height of the ball in Time T/2 is nh/4. Find n.

Given:-

• a ball is released from height h and it reaches ground in T.

• The acceleration of ball will be g.

• and initial velocity (u) will be zero.

To Find:-

we need to Find n.

Formula:-

$</p><p>\begin{gathered}\large \bf \red \bigstar \: \: \orange{ \underbrace{ \underline{s=ut+\dfrac{1}{2}at^2}}} \\\end{gathered}$

• with s = h,

• u = 0,

• t = T,

• a = g.

we have,

$h = \frac{1}{2} \: gT {}^{2} = > T {}^{2} = \frac{2h}{g} \: \: - - - > 1$

$And \: at \: t = \frac{T}{2} ,we \: have$

${\sf{{h_{1}}}}= \frac{1}{2} g \: ( \frac{T}{2} )^2 = \frac{1}{2} g \: \frac{T {}^{2} }{4}$

$=> \frac{g{\sf{{h_{1}}}}}{g} =T {}^{2} \: \: - - - > 2$

From Equation 1 and 2,

$\frac{g{\sf{{h_{1}}}}}{g} = \frac{2h}{g}$

$《But \: {\sf{{h_{1}}}} \: is \: from \: top \: of \: tower.》$

So,the ball of the height from the ground is

$h - \frac{h}{4} = \frac{3h}{4} = n = 3$

♤Answer:- n = 3