The body is fired with an initial angle theta0 and velocity v0. Show that the minimum time that it will reach the line AB is only when theta0=1/tan(alpha), where alpha is tha angle between the line AB and the horizontal.
To show that the minimum time for the body to reach the line AB occurs when theta0 equals 1/tan(alpha), we can use principles of projectile motion and calculus.
Let's assume that line AB is a horizontal line at some height h above the ground. We want to find the launch angle theta0 that minimizes the time it takes for the body to reach line AB.
1. First, let's break the initial velocity v0 into its horizontal and vertical components. The horizontal component (v0x) remains constant throughout the motion, while the vertical component (v0y) changes due to the acceleration of gravity.
2. The horizontal motion of the body is independent of the vertical motion. Therefore, we can focus on the vertical motion alone.
3. The equation for the vertical motion of an object under constant acceleration (due to gravity) is given by:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
- y is the vertical displacement from the initial height (y0),
- v0y is the initial vertical velocity,
- g is the acceleration due to gravity (9.8 m/s^2),
- t is the time.
4. Our objective is to find the time (t) it takes for the body to reach line AB, which means its vertical displacement (y) should be equal to the height of AB (h).
y = h
5. Substitute the expression for y in terms of t:
h = y0 + v0y*t - (1/2)*g*t^2
6. Since the body is launched with an initial angle theta0, we can express the initial vertical velocity (v0y) and the initial height (y0) in terms of theta0:
v0y = v0*sin(theta0)
y0 = 0
Substituting these values in the equation, we get:
h = v0*sin(theta0)*t - (1/2)*g*t^2
7. Now, let's differentiate the equation with respect to time (t) to find the time that minimizes the height (h):
dh/dt = v0*sin(theta0) - g*t
8. Set the derivative equal to zero to find the time (t) at which the body reaches the minimum height:
v0*sin(theta0) - g*t = 0
9. Solve for t:
t = v0*sin(theta0)/g
10. We know that the horizontal distance (x) traveled by the projectile in time t is given by:
x = v0x*t
= v0*cos(theta0)*t
11. The horizontal distance x is equal to the horizontal distance from the starting point to line AB. Let's call this distance d.
x = d
12. Substitute the expression for t that we obtained earlier:
d = v0*cos(theta0)*(v0*sin(theta0)/g)
13. Simplify the equation:
d = (v0^2*sin(theta0)*cos(theta0))/g
14. Using the trigonometric identity sin(2theta) = 2*sin(theta)*cos(theta), we can rewrite the equation as:
d = (v0^2*sin(2*theta0))/(2*g)
15. Since sin(2*theta0) has a maximum value of 1, to minimize the distance (d), sin(2*theta0) should also be at its maximum value.
sin(2*theta0) = 1
16. Rearranging the equation, we obtain:
2*theta0 = arcsin(1)
2*theta0 = pi/2
theta0 = pi/4
17. Finally, we know that alpha is the angle between line AB and the horizontal, so alpha = pi/4.
theta0 = 1/tan(alpha)
Therefore, the minimum time it takes for the body to reach line AB is when theta0 equals 1/tan(alpha), where alpha is the angle between line AB and the horizontal.