A fireworks rocket explodes at height h, the peak of its vertical trajectory. It throws out burning fragments in all directions, but all at the same speed v. Pellets of solidified metal fall to the ground without air resistance. Find the smallest angle that the final velocity of an impacting fragment makes with the horizontal. (Answer using g, h, and v, as needed.)
angle= tan-1 ((2gh)^1/2/v)
A firework rocket explodes at height h the peak of its vertical trajectory
To find the smallest angle that the final velocity of an impacting fragment makes with the horizontal, we can use vector components.
Let's consider the vertical motion of the fragment. The time it takes for the fragment to reach its peak and fall back down is the same, so we can use the equation for vertical motion:
h = (1/2) * g * t^2
Where:
h is the height
g is the acceleration due to gravity
t is the time taken
Simplifying the equation, we get:
t = √(2h/g)
Now, let's consider the horizontal motion of the fragment. The horizontal distance traveled by the fragment is the same as the distance traveled by a projectile launched horizontally with the same initial velocity v. The time taken for the horizontal motion is the same as the time taken for the vertical motion, which we found to be √(2h/g).
Using the equation for horizontal motion:
d = v * t
Where:
d is the horizontal distance
v is the initial velocity
t is the time taken
Substituting the value of t, we get:
d = v * √(2h/g)
Now, we can find the velocity of the fragment just before impact by calculating the magnitude of the vector combining the horizontal and vertical components of velocity:
v_final = √(v^2 + (2g * h))
Finally, to find the angle θ that the final velocity makes with the horizontal, we can use the inverse tangent function:
θ = arctan(v / √(2g * h))
Therefore, the smallest angle that the final velocity of an impacting fragment makes with the horizontal is arctan(v / √(2g * h)).
To find the smallest angle that the final velocity of an impacting fragment makes with the horizontal, we need to break down the problem into components and consider the projectile motion.
Let's assume that the initial velocity of the fragment is zero. The only force acting on the fragment is gravity, which causes it to accelerate downwards at a constant rate, denoted as 'g' (the acceleration due to gravity). The projectile motion occurs in a vertical plane.
When the fragment reaches the ground, its final velocity will have two components: the horizontal component (Vx) and the vertical component (Vy).
Since the initial velocity of the fragment is zero, its final velocity in the horizontal direction (Vx) remains zero throughout the motion. This is because there is no acceleration in the horizontal direction, as there are no horizontal forces acting on the fragment.
In the vertical direction, the final velocity (Vy) can be determined using the equation:
Vy^2 = Vo^2 + 2gh
Here, Vo refers to the initial velocity in the vertical direction, which is zero, and h represents the height from which the fragment is dropped.
Simplifying the equation, we have:
Vy^2 = 2gh
Taking the square root of both sides of the equation, we get:
Vy = √(2gh)
Now, we can find the tangent of the angle (θ) made by the final velocity of the fragment with the horizontal. The tangent of an angle is equal to the ratio of the vertical component of velocity to the horizontal component of velocity.
tan(θ) = Vy / Vx
Since Vx is zero, we can simplify the equation further:
tan(θ) = Vy / 0
Since Vy is non-zero, any division by zero is undefined. However, as the slope of a vertical line approaches infinity, the angle it forms with the horizontal approaches 90 degrees (or pi/2 radians).
Therefore, the smallest angle that the final velocity of an impacting fragment makes with the horizontal is 90 degrees (or pi/2 radians).