Impatient Sue wants to do Brilliant problems

Note that, the horizontal component of the velocity of the plane will make it reach 3000 meters above.

Now, we may calculate that, $300\;km\;per\;hour=\frac{300\times1000}{60\times60}=\frac{250}{3}m/s$

So, the horizontal component is, $V_y=\frac{250}{3}\sin30^\circ=\frac{125}{3}m/s$ [ $\sin30^\circ=\frac{1}{2}$ ]

The time to reach 3000 meters is, $t=\frac{3000}{V_y}=\frac{3000}{125}\times3=72\;seconds$

That much patience Sue needs to have.

This problem combines two major concepts: trigonometry and unit conversion, one after the other.

So first, the trigonometry. The best way to solve this problem is to draw a diagram, like so: https://i.imgur.com/KqoXDsl.png

So, the key that we need here is to see that the problem is asking for how long the VERTICAL ascent will take. We're given a diagonal ascent, and we need to get the vertical component out of that. This is done using the SIN of the degree:

sin(theta) = opposite/hypotenuse = vertical/diagonal sin(30 degrees) = vertical/300 Km/h vertical = sin(30 degrees) * 300 Km/h vertical = (1/2) * 300 Km/h vertical = 150 Km/h

So now we've finished the trigonometry. So, onto the unit conversion! Note that the problem is asking how many SECONDS it takes to reach 3000 METERS. Our current rate is in KILOMETERS per HOUR

The unit conversion is performed as follows:

First, we convert the hours to seconds (this is just my preference, it doesn't matter the order we do it in):

(150 Km/ 1 H) * (1 H / 60 M) * (1 M / 60 S) = 150 Km / (60 * 60 seconds) = 1/24 Km/s

Now, the Km to m conversion:

(1 Km / 24 s) * (1000 m / 1 Km) = 1000 m / 24 s = 125/3 m/s

So now, our final step: to take this rate that we've got, and figure out how long in seconds it takes to reach 3000 meters. This is the simplest part, and is performed as follows:

3000 m / (125/3 m/s) = 9000 m / 125 m/s = 72 s * Note here that the m on the bottom cancels with the m on the top, leaving us with just a second value, which is what the problem wanted.

Viola. 72 seconds.

really i cant understand one thing .... y is it given in "Details and Assumptions" tat typical angle of climb is 10-15 degrees wen v get the answer using sin 30 degrees????

we have to consider the vertical speed only.So 300km/h=300*1000/3600mtr/secnow mwe will multiply this with sin30.so we will get the vrtical speed.now we will devide the height 3000mtr by the above vertical component of speed and will get the answer that is 72 sec time required.

Given:

Angle of elevation = 30∘ Speed = 300 km/hr Altitude = 3000 m = 3 km

solution :

So, Pythagorean Theorem states that, in 30-60-90 degrees triangle, the hypotenuse is twice as long as the shorter leg .The altitude which is the shorter leg is 3 km so therefore, the distance or the hypotenuse is 3x2 = 6 km .

So, time = distance/speed >> 6 km/(300 km/hr ) = 0.02 hr

1 hr = 3600 sec >> 0.02 x 3600 = 72 seconds

SPEED=DISTANCE/TIME BUT THERE ISA CLIMB SO 300 KM/HRX SIN 30(1/20)=150 KM/HR SO EQUATE: 150 KM/3600 SECS=3 KM/X 150X=10800 X=72

sin 30∘=opposite side/hypotenuse

sin 30∘=3000m/hypotenuse

1/2=3000m/hypotenuse

hypotenuse=6000m or 6km

speed=distance/time

time=distance/speed

time=6km/300kph

time=0.02hr

time=0.02hr x 3600s/1hr=72s

d= (3000meters/sin30)x(1hour/300x10^3 meters)x(3600sec/hr) = 72 seconds

units cancel out... uhm...sin is opp/hyp... so opp is 3000....d is hyp :)))

The angle of climb of the airplane is 30 degrees which forms a special right triangle. The elevation of the plane is the shortest side and the distance is the hypotenuse. The plane must reach 3000 meters. 3000 multiplied by 2 is 6000. The plane must travel 6000 meters or 6 kilometers. It has a speed of 300 km an hour, and an hour is 3600 seconds long.

300/3600 * 6/x = (300x)=(21600) 21600/300=72

x=72

THE HEIGHT IS GIVEN WHICH IS 3 KM AND THE ANGLE RESPECT TO THE HORIZONTAL.THE HYPO IS 300T CAME FROM THE FORMULA RATE=DISTANCE/TIME.BY USING TRIGONOMETRY SIN 30X300T=3 . TIME IS 0.02 HRS CONVERT TO 72 SECS.

3000 m should be the planes altitude to access internet, the angle of elevation is 30° which means 3000 m is found opposite of 30°, it forms a special right triangle 30°-60°-90° where hypotenuse is twice the length of shortest leg so 3000m X 2 = 6000 m Then the distance formula... D = RT 0.6 km = (300 km/h)T T = 1/50 hr or 72 seconds

(300km/h 1000m)/(60 60)=1000/12 m/s. (changing speed from km/h to m/s) sin 30= 3000/distance. distance=3000 / sin30 = 6000 metres (using trig to find the distance travelled by the plane) time=6000/(1000/12)=72sec

sin 30 = h/s; 1/2 = 3000/s; s=6000; m=6km

t=s/v = 6/300 = 1/50 hour = 1/50x3600second = 72

velocity of the plane is = 300 km/h = 250/3 = 83.3333333333333 m/s &given altitude 3000 meter that mean the distance taken by the plane = 3000/sin(30deg)=6000 m. then time taken = 6000/83.3333 = 72 sec.

by diagram, the distance covered by the airplane is 3000/sin 30. time =(6*3600)/300=72

the displacement for the plane to reach 3000 meters high is 3000/(sin30)=6000 simply apply the formula s=vt, then you will get t=72s

Construct a triangle to solve this problem. From the problem description, you know that the angle that the plane climbs is 30 ^ /circ , so that can be the angle underneath your plane, which will climb your hypotenuse. 3000 meters, the desired height, is then your triangle's height. To find the length of the hypotenuse that the plane will have to travel to get to 3000 meters, use \sin 30^\circ , as this is your height divided by your hypotenuse, or \frac {3000 m}{Hypotenuse} . So, the distance the plane must travel to reach 3000 meters is 6000 meters, as the \sin 30^\circ = .5 . Then convert your 300 kilometers per second too meters per second and divide your hypotenuse to find the time it takes the plane in seconds.

300 . sen 30º(1/2)= 150 km/h 150/3,6=41,7m/s 3000m/41.7= 72s

The main mistake here for people is they will miss the kilometers part of the question. you have you multiply 300 by 1000 and then divide it by 6000 which gives you * (1/50 x 60* = 0.02 x 60 = 1.2 x 60 = 72)* the answer is 72

300km/hr =300000/60 m/m = 5000m/m = 5000/60 m/s =83.33m/s .Angle of elevation is 30 degree Hence distance traveled in 1 sec = 83.33 / 2 m (30-60-90 Theorm) For 3000m it will be 3000*2/83.33 = 72 sec

velocity in vertical direction = 300sin(30), v=displacement/time. now we know velocity in vertical direction displacement = 3 km . and by formula we get time =72

You can form a 30-60-80 degree triangle, in which , opposite of 30 degrees is 3000m or 3km. So the hypotenuse = 2(3km) = 6km. The distance , the plane must travel. So, 3km/300kph = .02hr = 72seconds.. :D

Let the taking off action of the airplane as a Pythagorean triangle with angle 30 degree ,the proper height of the plane to access Wifi is 3 kilometer which is opposite to the 30 degree angle ,so the Hypotenuse is 3*2=6 kilometer so the time to pass this distance is 72 sec after converting