A rocket fired at 275 m/s at 36o angle of elevation is described by:
x= (275cos〖36〗^0 )t and y= (275sin〖36〗^o )t-4.9t^2
Find the velocity (magnitude and direction) at t = 8 s.
Round to 1 decimal place.
To find the velocity (magnitude and direction) at t = 8 s, we need to differentiate the x and y equations with respect to time (t).
First, let's differentiate the x equation:
dx/dt = -275sin(36°)t
Next, let's differentiate the y equation:
dy/dt = 275cos(36°)t - 9.8t
Now, we have the components of velocity, dx/dt and dy/dt, at any given time t.
To find the velocity magnitude and direction at t = 8 s, we need to substitute t = 8 into the expressions we obtained.
Substituting t = 8 into dx/dt:
dx/dt = -275sin(36°)(8) = -1760.6 m/s
Substituting t = 8 into dy/dt:
dy/dt = 275cos(36°)(8) - 9.8(8) ≈ 1767.3 - 78.4 = 1688.9 m/s
Now, we have the components of velocity, dx/dt = -1760.6 m/s and dy/dt ≈ 1688.9 m/s, at t = 8 s.
To find the magnitude of velocity, we use the Pythagorean theorem:
|v| = sqrt((dx/dt)^2 + (dy/dt)^2)
|v| = sqrt((-1760.6)^2 + (1688.9)^2) ≈ 2366.1 m/s
Thus, the magnitude of velocity at t = 8 s is approximately 2366.1 m/s.
To find the direction of velocity, we use the inverse tangent function (arctan):
θ = arctan(dy/dt / dx/dt)
θ = arctan(1688.9 / -1760.6) ≈ -44.2°
Therefore, the direction of velocity at t = 8 s is approximately -44.2° (counterclockwise from the positive x-axis).