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°
it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram
Explanation:
Given:
v₀ = 22 m/s
v = 0 m/s
t = 17.32 s
Find: a
v = at + v₀
(0 m/s) = a (17.32 s) + (22 m/s)
a = -1.270 m/s²
Round as needed.
What is that?? Please tell us