The speed of a projectile when it reaches its
maximum height is 1/5 the speed when the projectile is at 1/5
its maximum height.
What is the initial projection angle?
vertical problem first, initial speed = Vi = S sin theta
what is the speed 1/5 of the way up
starting speed up = Vi
starting energy = 1/2 m Vi^2 + m g*0
energy at top = m g H = 1/2 m Vi^2
energy at 1/5 = m g H/5 + (1/2) m v^2 = 1/2 m Vi^2
so
(1/2) m Vi^2- (1/2) m v^2 = m g H/5
5 v^2 = 5 Vi^2 - 2 g H
but 2 g H = Vi^2
5 v^2 =4 Vi^2
v = Vi sqrt(4/5)
now speed at top = S cos theta
speed at 1/5 = sqrt [ (S cos theta)^2 + (S sin theta * sqrt4/5)^2]
so
1/5 = S cos theta / sqrt [ (S cos theta)^2 + (S sin theta * sqrt4/5)^2]
5 cos theta = sqrt [ ( cos theta)^2 + ( sin theta * sqrt4/5)^2]
5 cos theta = sqrt (cos^2 theta + 0.8 sin^2 theta )
24 cos^2 theta = 0.8 sin^2 theta
tan^2 theta = 30
tan theta = 5.48
theta = 79.6 deg
I did that fast, check the math!!!
To determine the initial projection angle, we need to analyze the motion of the projectile. Let's assume that the initial speed of the projectile is denoted by "V" and the angle of projection is denoted by "θ".
When a projectile is launched at an angle, its motion can be broken down into horizontal and vertical components. The horizontal component is unaffected by gravity and remains constant throughout the motion. The vertical component, however, is influenced by gravity.
The first piece of information we can extract from the problem is that the speed of the projectile when it reaches its maximum height is 1/5 the speed when it is at 1/5 of its maximum height. Let's denote these speeds as "V_max" and "V_1/5", respectively.
We know that the vertical component of the velocity at the maximum height is zero as the projectile momentarily stops before reversing direction. Therefore, the horizontal component of velocity remains constant, and the entire velocity is vertical at 1/5 the maximum height.
Now, let's break down the velocity components at these two points:
At the maximum height:
- Vertical component: 0 m/s
- Horizontal component: V_max * cos(θ)
At 1/5 the maximum height:
- Vertical component: V_1/5 * sin(θ)
- Horizontal component: V_1/5 * cos(θ)
Since the vertical component of the velocity is 1/5 of the maximum height when the projectile is at 1/5 its maximum height, we can write the following equation:
V_1/5 * sin(θ) = (1/5) * V_max
Dividing both sides by V_max, we get:
(V_1/5 / V_max) * sin(θ) = 1/5
Next, we need to relate the horizontal components of velocity at these two points. Since the horizontal component remains constant throughout the motion, we have:
V_max * cos(θ) = V_1/5 * cos(θ)
Canceling out the cos(θ) terms, we are left with:
V_max = V_1/5
Therefore, we can substitute V_max in the equation above to get:
(V_1/5 / V_max) * sin(θ) = 1/5
Simplifying further:
(sin(θ) / 1) = 1/5
This implies:
sin(θ) = 1/5
To find the angle θ, we can take the inverse sine (arcsin or sin^(-1)) of both sides:
θ = arcsin(1/5)
Using a calculator, the arcsin(1/5) is approximately 11.5 degrees.
So, the initial projection angle is approximately 11.5 degrees.