Compute first for the vertical motion, the formula is:
y = gt²/2
0.810 m = (9.81 m/s²)(t)²/2
t = 0.4064 s
whereas the horizontal motion is computed by:
x = (vx)t
4.65 m = (vx)(0.4064 s)
4.65 m/ 0.4064s = (vx)
(vx) = 11.44 m / s
So look for the final vertical speed.
(vy) = gt
(vy) = (9.81 m/s²)(0.4064 s)
(vy) = 3.99 m/s
speed with which it hit the ground:
v = sqrt[(vx)² + (vy)²]
v = sqrt[(11.44 m/s)² + (3.99 m/s)²]
v = 12.12 m / s
Explanation:
option A is the correct answer, if the gravitational acceleration is taken 10m/s²(rounding of 9.8/ms²).
hope this helps you.
F=ma
F= 4x1.2
F= 4.8 N
F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N
Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44
coefficient is 0.44
Let 
be the speed of the helicopter in still air. Let 
be the speed of the wind. Then, from the given information,
Adding the above 2 equations,
The speed of the helicopter in still air 
