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3 years ago

6

Contrary to popular belief, a ski jumper does not achieve a large amount of "air" when doing a jump (less than 6 feet). This is

because the ramp is almost horizontal while the landing is a steep hill. If a ski jumper leaves the hill at 26 m/s in the horizontal direction then travels 240 meters down the slope, what was the vertical distance (dy) that the ski jumper fell?

1 answer:

Elina [12.6K]3 years ago

6 0

Answer:

The vertical distance that the ski jumper fell is 417.45 m.

Explanation:

Given;

initial horizontal velocity of the jumper, $V_x$ = 26 m/s

horizontal distance of the jumper, dx = 240 m

The time of the motion is given by;

dx = Vₓt

t = dx / Vₓ

t = 240 / 26

t = 9.23 s

The vertical distance traveled by the diver is given by;

$d_y = V_yt + \frac{1}{2}gt^2$

initial vertical velocity, $V_y$ , = 0

$d_y = \frac{1}{2}gt^2\\\\d_y = \frac{1}{2}(9.8)(9.23)^2\\\\d_y = 417.45 \ m$

Therefore, the vertical distance that the ski jumper fell is 417.45 m.

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A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

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3 years ago

The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21

frutty [35]

Answer:

a) The velocity of the projectile at 2 seconds after launch is 1.9 meters per second. The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) The projectile reaches maximum height 2.192 seconds after launch.

c) The maximum height of the projectile is 26.584 meters above ground.

d) The projectile will hit the ground at 4.523 seconds after launch.

e) The velocity of the projectile right before hitting the ground in -22.871 meters per second.

Explanation:

Complete statement of problem is: <em>The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 21.5 meters per second is </em> $h(t) = 3+21.5\cdot t-4.9\cdot t^{2}$ <em>after t seconds. (Round your answers to two decimal places.) </em><em>(a)</em><em> Find the velocity after 2 seconds and after 4 seconds, </em><em>(b)</em><em> When does the projectile reach its maximum height? </em><em>(c)</em><em> What is the maximum height? </em><em>(d)</em><em> When does it hit the ground? </em><em>(e)</em><em> With what velocity does it hits the ground?</em>

a) From Physics and Differential Calculus we remember that velocity is the first derivative of height. Hence, we need to differentiate the height function in time:

$v(t) = 21.5-9.8\cdot t$ (Eq. 1)

Where $v(t)$ is the velocity function, measured in meters per second.

Now we evaluate this function at given times:

t = 2 s.

$v(2) = 21.5-9.8\cdot (2)$

$v(2) = 1.9\,\frac{m}{s}$

The velocity of the projectile at 2 seconds after launch is 1.9 meters per second.

t = 4 s.

$v(4) = 21.5-9.8\cdot (4)$

$v(4) = -17.7\,\frac{m}{s}$

The velocity of the projectile at 4 seconds after launch is -17.7 meters per second.

b) Maximum height is reached when velocity of projectile is zero. We equalize velocity to zero and solve the expression for $t$ :

$21.5-9.81\cdot t = 0$

$t = 2.192\,s$

The projectile reaches maximum height 2.192 seconds after launch.

c) Maximum height is calculated by evaluating height function at the time found in b). That is:

$h(2.192) = 3+21.5\cdot (2.192)-4.9\cdot (2.192)^{2}$

$h (2.192) = 26.584\,m$

The maximum height of the projectile is 26.584 meters above ground.

d) In this case, we need to equalize the height function to zero and solve for $t$ . That is:

$3+21.5\cdot t-4.9\cdot t^{2} = 0$

Roots are found by means of Quadratic Formula:

$t_{1}\approx 4.523\,s$ and $t_{2}\approx -0.135\,s$

Only the first root offers a physically reasonable solution. Therefore, the projectile will hit the ground at 4.523 seconds after launch.

e) This can be found by evaluating velocity function at the time found in d):

$v(4.523) = 21.5-9.81\cdot (4.523)$

$v(4.523) = -22.871\,\frac{m}{s}$

The velocity of the projectile right before hitting the ground in -22.871 meters per second.

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4 years ago

Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on

VikaD [51]

Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect

system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is

reversed. As a result, the velocity of the insect changes to a great amount. On the other hand,

the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect

suffers a greater change in momentum as compared to the car is correct. The momentum of

the insect after collision becomes very high because the car is moving at a high speed.

Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to

the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces

caused by the Newton’s action-reaction law. But, he made an incorrect statement as the

system suffers a change in momentum because the momentum before the collision is equal to

the momentum after the collision.

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3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

Ira Lisetskai [31]

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?

Solving:
$Ep = \frac{1}{2}*k*x^2$
$5184 = \frac{1}{2}*16200*x^2$
$5184*2 = 16200x^2$
$10368 = 16200x^2$
$16200x^2 = 10368$
$x^2 = \frac{10638}{16200}$
$x^2 = 0.64$
$x = \sqrt{0.64}$
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4 years ago

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