Question

A stone is thrown from a bridge at an angle of 30 down with the horizontal with a velocity of 25 m/s . If the stone strikes the water after 2.5 second then calculate the height of the bridge from the water surface ?​

Answer 1

AnSwer :

• Height of the bridge from the surface is 61.875 m.

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GiVen :

• A stone is thrown from a bridge at an angle of 30°.
• Velocity = 25 m/s
• Time at which stone strikes the surface = 2.5 s

To Find :

• Height of the bridge from the surface = ?

SoluTion :

Let the height of the bridge from surface be h.

★ Velocity :

$\implies$ - 25 sin 30

$\implies$ -25 /2

$\implies$ -12.5 m/s

By using the equations of motion,

$\implies$ S = ut + 1/2 at²

$\implies$ -h = - 12.5 x 2.5 - 1/2 x 9.8 x 2.5²

$\implies$ -h = - 31.25 - 30.625

$\implies$ h = 61.875 m

Therefore, the height of the bridge from the surface is 61.875 m.

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Answer 2

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$\huge\bold{\sf{\underline{\underline{\pink{QUESTION}}}}}$

➡ A stone is thrown from a bridge at an angle of 30 down with the horizontal with a velocity of 25 m/s . If the stone strikes the water after 2.5 second then calculate the height of the bridge from the water surface ?

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$\huge\bold{\sf{\underline{\underline{\pink{ANSWER}}}}}$

✤ CONCEPT AND FORMULA USED –

(Refer to the attachment)

❥ Here , the vertical component will be taken because it is the only component contributing in the speed of stone .

• It made two components - 25cos30° and 25sin30° .

• Only vertical component ( 25sin30° ) will be taken .

❥ FORMULA : $\sf s\:=\:ut\:+\: \dfrac{1}{2}a{t}^{2}$

${\red\bigstar\large\bold{\mathrm{\underline{\green{SØLUTION:}}}}}$

• Let h be the height attained .

➡ $\sf h\:=\:25sin30°(2.5)\:+\: \dfrac{1}{2}(9.8){2.5}^{2}$

➡ $\sf h\:=\:12.5 \times 2.5\:+\: 30.625$

➡ $\sf h\:=\:31.25\:+\: 30.625$

➡ $\sf h\:=\: 61.875$

${\red\bigstar\large\bold{\mathrm{\underline{\green{Height\:=\:61.9\:m}}}}}$

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