PHYSICS HIGH SCHOOL

A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the table of height h2, landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1+h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.A) 1) Without using equations, explain why making h1 very small would cause d to be small, even though

h2 would be very large?
2) Without using equations, explain why making h2 very small would cause d to be small, even though
h1 would be large

B) Derive an equation for d in terms of h1, h2, m, and physical constants as appropriate.

Answers

Answer 1

Answer:

(A)

(1) The reason for making $h_(1)$ very small is due to smaller value of horizontal component of launch velocity.

(2) The reason for making $h_(2)$ very small is due to smaller value of time of flight.

(B) The distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Given data:

The mass of block is, m.

The height of table from the top of slide is, $h_(1)$ .

The height of table at the end of slide is, $h_(2)$ .

The height of room is, H.

(A)

(1)

If the launch velocity of the block is v, then its horizontal component is very small, due to which adjusting the height $h_(1)$ to be very small will cause the d to be small.

(2)

The height $h_(2)$ is dependent on the time of flight, and since the time of flight taken by the block to get to the floor is very less, therefore the block will not get sufficient time to accomplish its horizontal motion. That is why making $h_(2)$ very small will cause d to be smaller.

(B)

The expression for the distance covered by the block is,

$v=(d)/(t)\nd = v * t$ ..............................(1)

Here, v is the launch speed of block and t is the time of flight.

The launch speed is,

$v^(2)=u^2+2gh_(1)\nv=\sqrt{u^2+2gh_(1)}\nv=\sqrt{0^2+2gh_(1)}\nv=\sqrt{2gh_(1)}$

And the time of flight is,

$h_(2)=ut+(1)/(2)gt^(2)\nh_(2)=0 * t+(1)/(2)gt^(2)\nh_(2)=0+(1)/(2)gt^(2)\nt=\sqrt(2h_(2))/(g)$

Substituting the values in equation (1) as,

$d = v * t\nd = \sqrt{2gh_(1)}* \sqrt(2 h_(2))/(g)}\nd=2\sqrt{h_(1)h_(2)}$

Thus, the distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Learn more about the time of flight here:

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Answer 2

Answer:

A1) The reason why making h₁ very small would cause d to be small is; Because the horizontal component of the launch velocity would be very small.

A2) The reason why making h₂very small would cause d to be small is;

Because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) The equation for d in terms of h₁ and h₂ is;

d = 2√(h₁ × h₂)

A) 1) The reason why making h₁ very small would cause d to be small is because the horizontal component of the launch velocity would be very small.

A) 2) The reason why making h₂ very small would cause d to be small is because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) Formula for Launch Velocity is;

V = √(2gh₁)

h₁ was used because the top of the slide from where the student released the block has a height of h₁.

Also, the time it takes to fall which is time of flight is given by the formula;

t = √(2h₂/g)

h₂ was used because the height of the table the object is on before falling is h₂.

Now, we know that d is distance from edge of the table and formula for distance with respect to speed and time is;

distance = speed × time

Thus;

d = √(2gh₁) × √(2h₂/g)

g will cancel out and this simplifies to give;

d = 2√(h₁ × h₂)

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Answers

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Answers

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Answers

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