Question

construct a triangle pqr in which pq =5cm ,qr =6cm and angle q=70°​

Answer 1

Answer:

Answer:-

\small \bf \underline{Given:}

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:}

Tofind:

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

\small \bf \underline{Solution:}

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)200−h=0.5×g×t........(1)

The height travelled by the ball vertically thrown upwards

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

2

........(2)

Adding equation (1) and (2) we get

200 = 40t

\bf \implies \: t = \frac{200}{40}⟹t=

40

200

\bf \implies \: t = 5s⟹t=5s

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

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

2

\bf \implies \: h = 200 - 4.9 \times 25⟹h=200−4.9×25

\bf \implies \: h = 200 - 122.5⟹h=200−122.5

\bf \implies \: h = 77.5m⟹h=77.5m

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

@BengaliBeauty

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Answer 2

Answer:

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