Question

a ball is thrown vertically upwards with an initial velocity such that it can reach a maximum height of 15m. if ,at the same instancea stone is dropped from a height of 15 m, find the ratio of distances travelled by them when they cross each other

Answer 1

Given:-

• Maximum height to which the ball is thrown is 15 m.

• At the same time, a stone is dropped from the height of 15 m.

To find:-

The ratio of distances travelled by them when they cross each other.

Solution:-

Let initial velocity of ball is u.  

At maximum height ,final velocity (v) = 0

Using the formula $\bold{v^2-u^2=2as}$,

$u^2=v^2-2as\\u=\sqrt{0-2*10*15}\\u=\sqrt{2*10*15}$

Suppose the ball travelled h₁ and the stone travelled h₂.

Using $\bold{s = ut +\frac{1}{2} \\ at^2}$,

h₁ = ut - 1/2 gt²

h₁ = √(2×10×15 )t - 1/2 gt²                  ...(i)

−h₂  = 0×t − 1/2  gt²

h₂  = 1/2 gt²                                       ...(ii)

Adding (i) and (ii):

​h₁ + h₂ = 15 = $\sqrt{2*10*15}t-\frac{1}{2}gt^2 +\frac{1}{2}gt^2$

So, 15 = $\sqrt{2*10*15}t$                        ...(iii)  

$t^2=\frac{15}{20}$

$\frac{1}{2} gt^2=\frac{15}{4}$                                           ...(iv)

From equations 3 and 4,

h₁ = 45/4 and h₂ = 15/4

h₁/h₂ = (45/4)/(15/4)

h₁/h₂ = 45/15

h₁/h₂ = 3/1

h₁:h₂ = 3:1

∴So, the ratio of distances travelled by them when they cross each other is 3:1.

Answer 2

Answer:

Given:

Ball thrown with Initial velocity such that it reaches max height 15 m. At the same Instant, a stone is dropped from 15 m

To find:

Ratio of distance travelled by them at the time they cross.

Calculation:

Let the distance travelled by Upper ball be x ,.then distance travelled by lower ball be (15-x).

Initial velocity of lower ball be u

For lower ball :

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

$= > {0}^{2} = {u}^{2} - 2 \times (10) \times (15)$

$= > u = \sqrt{300}$

$= > u = 17.32 \: m {s}^{ - 1}$

We should know that time taken to reach the meeting point will be same

For upper ball :

$x = \frac{1}{2} g {t}^{2} ........(1)$

For Lower ball :

$(15 - x )= (17.32 \times t) - \frac{1}{2} g {t}^{2} ......(2)$

Adding the 2 Equations , we get :

$15 = 17.32 \times t$

$= > t = 0.866 \: sec$

Putting value of t in 1st equation

$x = \frac{1}{2} g {t}^{2}$

$= > x = \frac{1}{2} \times 10 \times {(0.866)}^{2}$

$= > x = 4.33m$

So distance travelled by lower ball

$= 15 - 4.33 = 10.67 \: m$

So ratio will be :

10.67 : 4.33

= 2.43 : 1

So final answer :

$\: \: \: \: \: \: \: \: \: \boxed{ \huge{ \red{ \bold{2.43 \: : \: 1}}}}$