1. ## Question help...

Blah, I can't do these two questions. Could somebody please explain how you do these types of questions? (And maybe even the answer? ;D)

Q1. An aeroplane takes off from the ground at an angle of 27 degrees, and it's average speed in the first 10 seconds is 200 km/h. What is the altitude of the plane at the end of this time?

and

Q2. An observer notices that an aeroplane flies directly overhead at an altitude of 10 000m. Two minutes later the aeroplane is at an angle of elevation of 27 degrees. Determine the speed of the aeroplane.

Thankyou!

2. Originally Posted by Hikaru
Blah, I can't do these two questions. Could somebody please explain how you do these types of questions? (And maybe even the answer? ;D)

Q1. An aeroplane takes off from the ground at an angle of 27 degrees, and it's average speed in the first 10 seconds is 200 km/h. What is the altitude of the plane at the end of this time?

and

Q2. An observer notices that an aeroplane flies directly overhead at an altitude of 10 000m. Two minutes later the aeroplane is at an angle of elevation of 27 degrees. Determine the speed of the aeroplane.

Thankyou!
Q1. It is moving at a speed of 200 km/h at an angle of 27 degrees above the horizontal and it has traveled for 10 s. So it has covered a distance of
$s = vt = (200~km/h)(10~s) = (55.5556~m/s)(10~s) = 555.5556~m$
(I have converted this to m, but you can leave it in km if you like.)

So consider a right triangle where the base angle is 27 degrees and the hypotenuse is 556 m (or so). How do you find the height of this triangle?

Q2. Sketch this picture and you'll find pretty much the reverse problem of #1. Here you want to find the horizontal distance the plane has traveled from a right triangle and then you have the distance traveled and the time so you can find the speed.

-Dan