Question

A ball is allowed to fall from the top of a tower 200m height. At the same instant ,another ball is thrown vertically upword from the bottom of the tower with a velocity of 40m/s when and where the two balls meet​

Answer 1

Answer:

The balls will meet at a height of 77.5 m after a time of 5 seconds.

Explanation:

Given:

• A ball is allowed to fall from the top of a tower of height 200 m
• Another ball is thrown vertically upwards with a height of 40 m/s

To Find:

• Height at which the balls will meet
• Time at which the balls will meet

Solution:

Let us assume that the balls meet a time t seconds and a distance of s m from the ground.

Therefore at the time of meeting,

Distance travelled by ball which was dropped down = (200 - s) m

Distance travelled by the ball thrown upward = s m

For the ball moving downwards,

By the second equation of motion we know that,

s = ut + 1/2 × a × t²

where s = distance travelled

u = initial velocity

a = g = acceleration due to gravity

t =  time taken

Substitute the data,

(200 - s) = 0 × t + 1/2 × 9.8 × t²

200 - s = 4.9 t²

s = 200 - 4.9 t²-----(1)

For the ball thrown upwards,

s = ut + 1/2 × at²

Here

a = -g ( since the direction of motion of the object is opposite to that of acceleration due to gravity.)

Substitute the data,

s = 40 × t + 1/2 × -9.8 × t²

s = 40t - 4.9t²-------(2)

Since LHS of equation 1 and 2 are equal, RHS must also be equal.

Hence,

200 - 4.9t²= 40t - 4.9t²

200 = 40 t

t = 200/40

t = 5 s

Hence the balls will meet after a time of 5 seconds.

Substitute the value of t in equation 1,

s = 200 - 4.9 × 5²

s = 200 - 122.5

s = 77.5 m

Hence the balls will meet at a height of 77.5 m.

Answer 2

Answer:-

$\small \bf \underline{Given:}$

★★ Height at which the ball is allowed to fall from the top of the tower = 200m

★★ Velocity at which the other ball is thrown vertically upwards from the bottom of the tower = 40 m/s

$\small \bf \underline{To \: find:}$

We need to find when and where does the two balls meet

$\small \bf \underline{Solution:}$

Let the ball meet at height h from the ground

Therefore, the dropped ball travels a height = 200-h

Distance travelled by the ball thrown vertically upward = h

Let t be the time at which they meet

The height travelled by the dropped ball

$\bf200 - h = 0.5 \times g \times t \: \: \: \: \: \: \: \: \: ........(1)$

The height travelled by the ball vertically thrown upwards

$\bf \: h = 40t - 0.5 \times g \times {t}^{2} \: \: \: \: \: \: \: \: ........(2)$

Adding equation (1) and (2) we get

200 = 40t

$\bf \implies \: t = \frac{200}{40}$

$\bf \implies \: t = 5s$

Substituting the value of t in equation (1), the equation stands as follows:

$\bf \: 200 - h = 4.9 \times {5}^{2}$

$\bf \implies \: h = 200 - 4.9 \times 25$

$\bf \implies \: h = 200 - 122.5$

$\bf \implies \: h = 77.5m$

Answer: The two balls meet at a height of 77.5m from the ground after 5s

@BengaliBeauty

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