Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
1.27 m
Explanation:
Distance = 192 m
number of rotations = 48
Distance traveled in one rotation = 2 x π x r
Where, r be the radius of wheel.
so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r
It is equal to the distance traveled.
192 = 48 x 2 x 3.14 x r
r = 0.637 m
diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m
The minimum is 3 Newton, when the two forces act in opposite directions.
Answer:
d = 380 feet
Explanation:
Height of man = perpendicular= 130 feet
Angle of depression = ∅ = 70 °
distance to bus stop from man = hypotenuse = d = 130 sec∅
As sec ∅ = 1 / cos∅
so d = 130 sec∅ or d = 130 / cos∅
d = 130 / cos(70°)
d = 380 feet
0.6 cm is the answer add it up and find the m/s hope this helps