Question

A wicketkeeper gloves dropped from a height of 40 m and simultaneously a ball is thrown upward from the ground with a speed of 40 m/s .when and where do they meet?

Answer 1

$\huge \mathfrak {Answer:-}$

Given:

Height $=40\:m$

Speed $= 40\:m/s$

By laws of motion, [Gloves downward motion]

Initial velocity $=0$

We get,

Height $= ut \:+$$\frac{1}{2}$$gt ^{2}$

$x = 0 + \frac{1}{2} \times 9.8 \times t ^{2}$

$x = 4.9t ^{2}$

Now,

Ball upward motion,

Initial velocity $=40$

Using,

$h = ut + \frac{1}{2} gt ^{2}$

We get,

$40 - x = 40xt + \frac{1}{2} \times 9.8 \times t^{2}$

$40 - x = 40t - 4.9t^{2}$

$40 - x = 40t + x$

$40 = 40t$

Where,

$t = 1\:second$

Therefore,

$x = 4.9 \times t^{2}$

$= 4.9 \times 1 ^{2}$

$= 4.9m$

Height $=4.9\:m$

Now,

The meeting height

$= 40 - 4.9$

$= 35.1m$

So,

The glove will meet the ball at $35.1m$ above the ground after $1\:second.$

$\huge {Be\:Brainly}$❤️

Answer 2

${ \huge{ \sf{ Answer :}}}$

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Given :-

Height : 40 metre

Speed : 40 m/s

To find :

When they meet : ?
Where they meet : ?

Solution :

Let's consider that the glove and the ball meet at time 't'

Let's consider that they meet at height 'H'

Now,

According to the formula ,

$\sf{H = ut + \frac{1}{2} gt^2}$

As u = 0,

$\sf{H = \frac{1}{2} gt^2}$

$\sf{H = \frac{1}{2} × 9.8 t^2 }$

$\sf{ H = 4.9 t^2}$

Let this be (1)

Now,

According to the next given condition,

u = 40 m/s

$\sf{h = ut + \frac{1}{2} gt^2}$

$\sf \tiny{ \therefore \: 40 - H = 40 \times t + \frac{1}{2} \times ( - 9.8)t {}^{2}}$

$\sf{40 - H = 40 \times t - 4.9t {}^{2}}$

Now , we know that , h = 4.9t^2,

Hence,

$\sf{40 - H = 40 \times t -H {}^{2}}$

Eliminating H,

$\sf{40 = 40t}$

$\sf{t = 1 \: sec}$

Hence,

h = 4.9t^2,

= 4.9 × 1^2

h = 4.9 metre

Also,

40 - 4.9 = Height from the ground

h = 35.1 metre above the ground

Final answer :-

After 1 second, the glove and the ball will meet.

They will meet at a distance of 35.1 metre above the ground.

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