Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
The amplitude is from the absolute value of the 0 point on the y-axis to the highest(peak) or lowest(troph) point of the wave. In this question, 3cm is the highest and -3cm is the lowest, so the amplitude is 3cm.
The dotted path is the path of the ball. it reaches it's maximum height at the top where vertical, y-velocity = 0
The initial y-velocity = 19sin(70°)
initial y-velocity = 17.85 m/s
Use one of the kinematic equations with velocity and time. No displacement because we don't want to worry about figuring that out.
v = u - gt
0 = 17.85 - 9.8t
-17.85 = -9.8t
17.85/9.8 = t
1.82 sec = t