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 a specific time (t=8s) using the given equations x=(275cos36°)t and y=(275sin36°)t-4.9t^2, we need to differentiate both equations with respect to time t to obtain the equations for velocity components.
1. Differentiating the equation x=(275cos36°)t with respect to t:
dx/dt = (275cos36°)
2. Differentiating the equation y=(275sin36°)t-4.9t^2 with respect to t:
dy/dt = (275sin36°) - 9.8t
Now we have the velocity components:
Vx = dx/dt = (275cos36°) (Equation 1)
Vy = dy/dt = (275sin36°) - 9.8t (Equation 2)
To find the magnitude and direction of the velocity, we can use the Pythagorean theorem:
V = √(Vx^2 + Vy^2)
Substituting the values from Equations 1 and 2:
V = √((275cos36°)^2 + ((275sin36°) - 9.8t)^2)
Now we can calculate the magnitude of the velocity at t = 8s:
V = √((275cos36°)^2 + ((275sin36°) - 9.8(8))^2)
Using a calculator, evaluate this expression and round to 1 decimal place to get the magnitude of the velocity at t = 8s.
To determine the direction of the velocity, we need to find the angle at which the velocity vector points. We can use trigonometric functions to calculate the angle:
θ = arctan(Vy / Vx)
Substituting the values from Equations 1 and 2:
θ = arctan(((275sin36°) - 9.8t) / (275cos36°))
Now we can calculate the direction of the velocity at t = 8s:
θ = arctan(((275sin36°) - 9.8(8)) / (275cos36°))
Using a calculator, evaluate this expression to get the direction of the velocity vector at t = 8s.