Question

A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case, and air resistance is negligible.Case A: Thrown straight up.Case B: Thrown straight down.Case C: Thrown out at an angle of 45 above horizontal.Case D: Thrown straight out horizontally.In which case will the speed of the stone be greatest when it hits the water below?

Answer 1

Answer:

the velocity in all the cases will be same.

Explanation:

given,

girl throws a stone from the bridge

air is friction less

we have to find from the given cases in which case the velocity of stone will be greatest.

According to Work energy theorem work done by the sum of all the force is equal to kinetic energy.

As the air is frictionless hence the speed depend upon the height from which the stone is thrown.

height in all the cases is same.

so, the velocity in all the cases will be same.

Answer 2

The speed of the stone hitting the water below will be the same for every case.

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Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

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Given:

height of the stone = h

initial speed of the stone = u

Unknown:

final speed of the stone = v = ?

Solution:

Case A:

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

$v^2 = u^2 - 2g(-h)$

$v^2 = u^2 + 2gh$

$\boxed {v = \sqrt{u^2 + 2gh}}$

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Case B:

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

$v^2 = (-u)^2 - 2g(-h)$

$v^2 = u^2 + 2gh$

$\boxed {v = \sqrt{u^2 + 2gh}}$

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Case C:

$v_x = u \cos \theta$

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$(v_y)^2 = (u \sin \theta)^2 - 2gh$

$(v_y)^2 = (u \sin \theta)^2 - 2g(-h)$

$(v_y)^2 = (u \sin \theta)^2 + 2gh$

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$v^2 = (v_x)^2 + (v_y)^2$

$v^2 = (u \cos \theta)^2 + (u \sin \theta)^2 + 2gh$

$v^2 = u^2 ( \cos^2 \theta + \sin^2 \theta) + 2gh$

$v^2 = u^2 (1) + 2gh$

$\boxed{v = \sqrt{u^2 + 2gh}}$

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Case D:

$v_x = u \cos \theta$

$v_x = u \cos 0^o$

$v_x = u$

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$(v_y)^2 = (u \sin \theta)^2 - 2gh$

$(v_y)^2 = (u \sin 0^o)^2 - 2g(-h)$

$(v_y)^2 = (0)^2 + 2gh$

$(v_y)^2 = 2gh$

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$v^2 = (v_x)^2 + (v_y)^2$

$v^2 = (u)^2 + 2gh$

$\boxed{v = \sqrt{u^2 + 2gh}}$

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From information above , we could conclude that the speed of the stone hitting the water below will be the same for every case.

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Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall