Given that,
For glasses:
distance, s = 42 m
a= -g = -9.81 m/s²
we can calculate time taken by glasses to hit the ground as follow:
s= Vi* t + 1/2 at²
Since at top initial velocity, Vi= 0
42 = 0 - 0.5* 9.81 t²
t = 2.926 s
Pen is drop 2 second after the glasses is dropped. So time difference will be:
Δt = 2.926 - 2 = 0.926 s
Distance covered by Pen in 0.926 s can be calculated as:
s = 0 - 0.5* 9.81 * 0.926²
s= 4.2 m
Distance of pen from ground will be, 42-4.2 = 37.8 m
When the glasses hit the ground, the pen will be 37.8 m above the ground.
Answer:
speed wind Vw = 54.04 km / h θ = 87.9º
Explanation:
We have a speed vector composition exercise
In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south
Let's add the vectors in each coordinate axis
X axis (East-West)
-Xvion - Xw = -119
Xw = -Xavion + 119
Xw = 119 -108
Xwi = 1 km
Calculate the speed for time of t = 0.5 h
Vwx = Xw / t
Vwx= 1 /0.5
Vwx = - 2 km / h
Y Axis (North-South)
Y plane - Yi = -27
Y plane = 0
Yw = 27 km
Vwy = 27 /0.5
Vwy = 54 km / h
Let's use the Pythagorean theorem and trigonometry to compose the answer
Vw = √ (Vwx² + Vwy²)
Vw = R 2² + 54²
Vw = 54.04 km / h
tan θ = Vwy / Vwx
tan θ = 54/2 = 27
θ = Tan⁻¹ 1 27
θ = 87.9º
The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º west from the south
The mass of a dwarf planet.