A projectile is launched at an angle of 60 degrees to the horizontal from 6.5 ft above the ground at an initial speed of 150ft/sec. Assume the x-axis is horizontal, the positive y-axis is vertical (opposite g), the ground is horizontal, and only the gravitational force acts on the object.
Find the velocity and position vectors for t ≥ 0?
ie. The velocity vector is v(t) = <__,__>
To find the velocity and position vectors for a projectile launched at an angle of 60 degrees, we can break down the motion into its horizontal and vertical components.
The initial speed of the projectile is given as 150 ft/sec, and the angle of projection is 60 degrees. This means that the initial velocity components can be calculated as follows:
Initial horizontal component of velocity (Vx0) = initial speed * cos(angle)
Vx0 = 150 ft/sec * cos(60 degrees)
Vx0 = 150 ft/sec * 0.5
Vx0 = 75 ft/sec
Initial vertical component of velocity (Vy0) = initial speed * sin(angle)
Vy0 = 150 ft/sec * sin(60 degrees)
Vy0 = 150 ft/sec * 0.866 (approximated)
Vy0 = 129.9 ft/sec (approximated)
Now we can express the velocity vector v(t) as:
v(t) = <Vx(t), Vy(t)>
For the horizontal component, the velocity remains constant throughout the motion as there is no horizontal acceleration. Therefore, Vx(t) = Vx0 = 75 ft/sec.
For the vertical component, the velocity changes due to the acceleration due to gravity. We can use the equation of motion to calculate the vertical velocity at any time t:
Vy(t) = Vy0 - g*t
where g is the acceleration due to gravity (approximated as 32.2 ft/sec^2) and t is the time since the projectile was launched.
Thus, the velocity vector v(t) is:
v(t) = <75 ft/sec, Vy0 - g*t>
Now, let's find the position vector r(t) which gives the position of the projectile at any time t.
The horizontal position of the projectile (x-coordinate) can be calculated using the equation:
x(t) = x0 + Vx0*t
where x0 is the initial horizontal position, which is 0 in this case.
Therefore, x(t) = 0 + 75 ft/sec * t = 75t ft.
The vertical position of the projectile (y-coordinate) can be calculated using the equation:
y(t) = y0 + Vy0*t - (0.5)*g*t^2
where y0 is the initial vertical position, which is 6.5 ft in this case.
Therefore, y(t) = 6.5 ft + (129.9 ft/sec * t) - (0.5) * (32.2 ft/sec^2) * (t^2).
Therefore, the position vector r(t) is:
r(t) = <x(t), y(t)>
r(t) = <75t ft, 6.5 + (129.9t ft/sec) - (0.5) * (32.2 ft/sec^2) * (t^2)>