Yeah!! It is better to decrease the speed of the vehicle so the ejection of gases would happen in slow manner.... then you can save your gasoline
Answer:Velocity = 6.325m/s
Directional angle= 18.43°
Explanation:
Using Right angle triangle
Let Velocity of ballon&hawk be VHB represent the height of the triangle.
Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.
Let Velocity of hawk and ground BE VHG represent the hypothesis.
Theta = opp/Adj= VHB/VBG
using pythagorean
VHG= SQRT(VHB^2+VBG^2)
VHG= sqrt(2^2+6^2)
VHG= sqrt(4+36)
VHG= 6.325m/s
Tan theta= 2/6
Tan theta =0.3333
Tan^-1 0.3333=18.43°
Answer:
31.75 m/s
Explanation:
h = 41.7 m
Let the initial velocity of the second stone is u
Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.
For first stone:
Use second equation of motion

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7
So, 41.7= 0 + 0.5 x 9.8 x t^2
41.7 = 4.9 t^2
t = 2.92 s ..... (1)
For second stone:
Use second equation of motion

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity
.... (2)
By equation the equation (1) and (2), we get

u = 31.75 m/s
its D
i hope this helps to future test takers. i took the test and had to retake it because their wasn't an answer