Answer:
[tex]V_0[/tex]
Explanation:
Given that, the range covered by the sphere, [tex]M[/tex], when released by the robot from the height, [tex]H[/tex], with the horizontal speed [tex]V_0[/tex] is [tex]D[/tex] as shown in the figure.
The initial velocity in the vertical direction is [tex]0[/tex].
Let [tex]g[/tex] be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. [tex]V_0[/tex] will remain constant throughout the projectile motion.
So, if the time of flight is [tex]t[/tex], then
[tex]D=V_0t\; \cdots (i)[/tex]
Now, from the equation of motion
[tex]s=ut+\frac 1 2 at^2\;\cdots (ii)[/tex]
Where [tex]s[/tex] is the displacement is the direction of force, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the constant acceleration and [tex]t[/tex] is time.
Here, [tex]s= -H, u=0,[/tex] and [tex]a=-g[/tex] (negative sign is for taking the sigh convention positive in [tex]+y[/tex] direction as shown in the figure.)
So, from equation (ii),
[tex]-H=0\times t +\frac 1 2 (-g)t^2[/tex]
[tex]\Rightarrow H=\frac 1 2 gt^2[/tex]
[tex]\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)[/tex]
Similarly, for the launched height [tex]2H[/tex], the new time of flight, [tex]t'[/tex], is
[tex]t'=\sqrt {\frac {4H}{g}}[/tex]
From equation (iii), we have
[tex]\Rightarrow t'=\sqrt 2 t\;\cdots (iv)[/tex]
Now, the spheres may be launched at speed [tex]V_0[/tex] or [tex]2V_0[/tex].
Let, the distance covered in the [tex]x-[/tex]direction be [tex]D_1[/tex] for [tex]V_0[/tex] and [tex]D_2[/tex] for [tex]2V_0[/tex], we have
[tex]D_1=V_0t'[/tex]
[tex]D_1=V_0\times \sqrt 2 t[/tex] [from equation (iv)]
[tex]\Rightarrow D_1=\sqrt 2 D[/tex] [from equation (i)]
[tex]\Rightarrow D_1=1.41 D[/tex] (approximately)
This is in the [tex]3[/tex] points range as given in the figure.
Similarly, [tex]D_2=2V_0t'[/tex]
[tex]D_2=2V_0\times \sqrt 2 t[/tex] [from equation (iv)]
[tex]\Rightarrow D_2=2\sqrt 2 D[/tex] [from equation (i)]
[tex]\Rightarrow D_2=2.82 D[/tex] (approximately)
This is out of range, so there is no point for [tex]2V_0[/tex].
Hence, students must choose the speed [tex]V_0[/tex] to launch the sphere to get the maximum number of points.
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
Learn more about free fall motion here:
https://brainly.com/question/17842118
