42The full set of question:
A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizontal direction with variable speed and from a variable vertical position and a fixed horizontal position x=0.
The robot is calibrated by adjusting the speed at which the sphere is launched and the height of the robot’s sphere launcher. Depending on where the spheres land on the ground, students earn points based on the accuracy of the robot. The robot is calibrated so that when the spheres are launched from a vertical position y=H and speed v0, they consistently land on the ground on a target that is at a position x=D. Positive directions for vector quantities are indicated in the figure.
When the students arrive at the competition, it is determined that the height of the sphere launcher can no longer be adjusted due to a mechanical malfunction. Therefore, the spheres must be launched at a vertical position of y=H2. However, the spheres may be launched at speed v0 or 2v0.
Question: In a clear response that may also contain diagrams and/or equations, describe which speed, v0 or 2v0, the students should launch the sphere at so that they earn the maximum number of points in the competition.
Answer:
Free fall motion is motion in which the only force acting on the body is gravity
The student should launch the sphere at 2·v₀,
for the sphere will land at approximately 1.41·D, which is in the 3 point zone
The given parameter are:
The distance covered by the sphere when launched at height, H = D
The velocity with which the ball reaches D = v₀
The current available height of launcher= H/2
The available velocities = v₀, and 2·v₀
Solution:
From H = u·t + (1/2)·g·t², where, initial velocity of the vertical motion of the ball, u = 0 we have;
H = (1/2)·g·t²
∴ The time it takes the ball to drop from H, t = √(2·H/g)
The distance, D = v₀ × √(2·H/g)
When the height is H/2, we get:
t = √(2·H/(2·g)) = √(H/g)
The distance covered, D₁ = v₀ × √(H/g)
Therefore, D = (√2) × v₀ × √(H/g) = (√(2))·D₁
D₁ = D/(√2) ≈ 0.71·D
D₁ ≈ 0.71·D
At speed 2·v₀, we have;
D₂ = 2·v₀ × √(H/g) = √2 × v2 × v₀ × √(H/g) = √2 × v₀ × √(2·H/g) = √2·D₁ ≈ 1.41·D
D₂ ≈ 1.41·D
The 2 point zone = D/2 < x < D = 0.5·D < x < D (Position D₁ ≈ 0.71·D is located here)
The 3 Point Zone = D < x < 3·D/2 = D < x < 1.5·D (Position D₂ ≈ 1.41·D is located here)
Given that at D₂, the ball lands in the 3 Point Zone, the student should launch the sphere at the speed, 2·v₀, so that the ball will land at D₂ ≈ 1.41·D, which is in the 3 Point Zone
p.s. Question (a) is not connected to this question at all. The right second part of this question is shown above.
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rating
, when released by the robot from the height, 
, with the horizontal speed 
as shown in the figure.
.
be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. 
, then
is the displacement is the direction of force, 
is the initial velocity, 
is the constant acceleration and 
and 
(negative sign is for taking the sigh convention positive in 
direction as shown in the figure.)
, the new time of flight, 
, is
.
direction be 
for 
for 
[from equation (iv)]
[from equation (i)]
(approximately)
points range as given in the figure.
[from equation (iv)]
[from equation (i)]
(approximately)
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