Knowing the initial velocity and angle, the horizontal range formula is given by R= V^2sin(2teta) / g, so we can get
sin(2teta)=Rg/V^2
sin(2teta)= (180 x 9.8)/ 80^2= 0.27, sin(2teta)=0.27, 2teta=arcsin(0.27)=15.66, so teta=15.66/2
teta=7.83°
Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.
Answer:
a) x = 8.8 cm * cos (9.52 rad/s * t)
b) x = 8.45 cm
Explanation:
This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.
From the question, we are given that
A = 8.8 cm = 0.088 m
t = 0.66 s
Now, we need to find the angular speed w, such that
w = 2π/T
w = (2 * 3.142) / 0.66
w = 6.284 / 0.66
w = 9.52 rad/s
The displacement equation of Simple Harmonic Motion is usually given as
x = A*sin(w*t)
But then, the equation starts from the equilibrium point at 0 sec, i.e x = 0 m
When you have to start from the max displacement, then the equation would be
x = A*cos(w*t).
So when t = 0 the cos(0) = 1, and then x = A which is max displacement.
Thus, the equation is
x = 8.8 cm * cos (9.52 rad/s * t)
At t = 1.7 s,
x = 8.8 cos (9.52 * 1.7)
x = 8.8 cos (16.184)
x = -8.45 cm
What following terms I don’t see any please make your question clear
Answer: A
Explanation: Any short-duration exercise that is powered primarily by metabolic pathways that do not use oxygen. Examples
of anaerobic exercise include sprinting and weight lifting.
