A rock is thrown from the roof of a building with a velocity v0 at an angle α0. The
height of the building is h.
(i)Calculate the horizontal distance traveled by the projectile before it strikes the
ground.
(ii)Show that the speed of the rock just before it strikes the ground is independent
of α0.
(iii)Calculate the value of the launching angle for maximum horizontal distance
when the building height is zero, h = 0.
To solve these questions, we will use the principles of projectile motion. We can consider the motion of the rock in two separate directions: vertical and horizontal. Let's break down the solution step by step for each question:
(i) Calculate the horizontal distance traveled by the projectile before it strikes the ground:
First, observe that the vertical component of the rock's initial velocity, v0sin(α0), determines how long it takes for the rock to reach its maximum height (where the vertical velocity becomes zero). The time taken, t, can be calculated using the equation:
h = (v0sin(α0))t - (1/2)gt^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Simplifying the equation, we get:
t = (v0sin(α0)) / g
Now, we know that the horizontal component of the initial velocity, v0cos(α0), remains constant throughout the motion. The horizontal distance traveled, d, can be calculated by multiplying the horizontal velocity by the time:
d = (v0cos(α0)) * t
(ii) Show that the speed of the rock just before it strikes the ground is independent of α0:
We can prove that the speed of the rock just before it hits the ground is independent of the launch angle, α0, by considering the final vertical velocity, vf, just before it strikes the ground.
Using the equations of motion, we know that vf = v0sin(α0) - gt. Since we are only interested in the magnitude of vf, we can drop the negative sign and write it as:
vf = v0sin(α0) + gt
Now, the time taken for the rock to reach the ground, t_total, is given by:
t_total = 2(v0sin(α0)) / g
Substituting this value of t_total into the velocity equation, we get:
vf = v0sin(α0) + g * (2(v0sin(α0)) / g)
vf = v0sin(α0) + 2v0sin(α0)
vf = 3v0sin(α0)
Notice that the final velocity, vf, is dependent on the initial velocity, v0, and the angle of projection, α0, but not the sign of α0. Therefore, the speed of the rock just before it hits the ground is independent of α0.
(iii) Calculate the value of the launching angle for maximum horizontal distance when the building height is zero, h = 0:
When the building height is zero, the maximum horizontal distance can be achieved by maximizing the range, d. To find the angle, α_max, that gives the maximum distance, we differentiate d with respect to α0, set it equal to zero, and solve for α0.
By differentiating d = (v0cos(α0)) * t with respect to α0, we get:
dd/α0 = (v0cos(α0)) * dt/α0
To find dt/α0, we need to differentiate the equation in part (i), h = (v0sin(α0))t - (1/2)gt^2, with respect to α0:
dh/dt = (v0sin(α0)) * dt/α0 - gt
Setting dh/dt equal to 0, we get:
0 = (v0sin(α0)) * dt/α0 - gt
Rearranging the equation, we have:
gt = (v0sin(α0)) * dt/α0
Substituting this result into dd/α0, we get:
dd/α0 = (v0cos(α0)) * gt / α0
Simplifying further, we have:
dd/α0 = (v0cos(α0)) * (v0sin(α0)) / α0
To maximize d, dd/α0 = 0. So, we solve the equation:
0 = (v0cos(α0)) * (v0sin(α0)) / α0
Since v0 is non-zero, we can cancel it out:
0 = cos(α0) * sin(α0) / α0
To simplify further, we can use the identity sin(2α) = 2sin(α)cos(α):
0 = sin(2α0) / α0
For the equation to hold true, sin(2α0) = 0, so 2α0 = 0, π, 2π, etc. Therefore, the value of α0 for maximum horizontal distance, when the building height is h = 0, is α0 = 0, π/2, π, or any other angle that satisfies sin(2α0) = 0.