Physics, 04.07.2019 05:30 KaleahV

Somewhere in the pacific, on a remote island, patrick was launched into the air because he was holding onto an icbm (rocket) which began rising at a constant velocity of 68.0 m/s, straight up. at an altitude of 1500.0 meters patrick lost his grip on the rocket. at this very instant, spongebob was 12.00 km away, sitting in his trusty airboat, jet engine idling of course, at an altitude of 1550.0 m. at what speed must spongebob travel, without deviating from his current altitude, to save his best friend and buddy?

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Answer from: bri2008

Spongebob will have to move at a speed of 15414 m/s to catch Patrick while he is rising or at a speed of 916.0 m/s to catch him as he falls down.

Patrick rises up to an altitude of 1500.0 m with a constant velocity, when he loses contact with the ICBM. His motion after he loses contact is similar to any body which is projected vertically upwards with an initial velocity u. He is then acted upon by acceleration due to gravity, which is directed vertically downwards, towards the center of the earth. Patrick rises up from the point at which he loses contact with the ICBM, to a vertical height, where his velocity becomes equal to zero, after which he falls down.

Spongebob is at a distance d away from the point of projection of Patrick at an altitude of 1550.0 m. The difference between the heights at which Patrick is projected and the height at which SpongeBob idles is s, which is given by,

s= 1550 m-1500m=50 m

He can catch Patrick at two different times-(i) while he rises up or (ii) when he falls down to the height s from his initial point of projection.

Use the equation of motion

$s=ut+\frac{1}{2} at^2$

Substitute +50 m for s, +68.0 m/s for u and -9.8 m/s² for g.

$s=ut+\frac{1}{2} at^2\\(50m)=(68.0m/s)t+\frac{1}{2} (-9.8m/s^2)t^2$

Solve the quadratic equation for t.

Taking the positive root,

$t=\frac{68+60.37}{9.8} =13.1 s$

taking the negative root,

$t=\frac{68-60.37}{9.8} =0.7785$

In the time of 0.7785 s, Patrick first reaches the height of 50 m from the point where he has lost contact and if SpongeBob has to catch him, he needs to go at a speed,

$v=\frac{12000 m}{0.7785s} =15414m/s$

In the time of 1.1 s, he reaches the point again, after dropping down from the maximum height reached. The speed SpongeBob needs to travel in this case is

$v=\frac{12000 m}{13.1s} =916.03m/s$

Spongebob will have to move at a speed of 15414 m/s to catch Patrick while he is rising or at a speed of 916.0 m/s to catch him as he falls down.

Answer from: Quest

waves moving from a fast medium to a slow medium will experience a reflection as light travels more slowly than red light. total internal reflection occurs when the critical angle is.

Answer from: Quest

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

time (seconds)

answer: 0 meters/second^2

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Somewhere in the pacific, on a remote island, patrick was launched into the air because he was holdi...