A cannon fires a projectile into the air at an angle of 60degrees above the horizontal. The projectile falls back on the ground at a distance d from the cannon and its maximum height is h. Ratio d/h is equal to:
To find the ratio d/h, we can make use of the equations of motion for projectile motion.
The horizontal motion of the projectile can be described by the equation:
d = v₀ * t * cos(θ)
where d is the horizontal distance traveled, v₀ is the initial velocity of the projectile, t is the total time of flight, and θ is the launch angle.
The vertical motion of the projectile can be described by the equation:
h = v₀ * t * sin(θ) - (1/2) * g * t²
where h is the maximum height reached and g is the acceleration due to gravity.
Given that the angle of launch is 60 degrees (θ = 60°), and assuming that the initial velocity v₀ is the same for both horizontal and vertical motions, we can rewrite the equation for h as:
h = (v₀² * sin²(θ)) / (2 * g)
Now, let's find the time of flight (t):
The time of flight, t, can be calculated using the equation:
t = 2 * (v₀ * sin(θ)) / g
Now, let's substitute this value of t into the equation for d:
d = v₀ * t * cos(θ)
= 2 * (v₀ * sin(θ)) / g * v₀ * cos(θ)
= (2 * v₀² * sin(θ) * cos(θ)) / g
Dividing d by h gives us the ratio:
d/h = [(2 * v₀² * sin(θ) * cos(θ)) / g] / [(v₀² * sin²(θ)) / (2 * g)]
= 4 * cos(θ) / sin(θ)
Since θ = 60°,
d/h = 4 * cos(60°) / sin(60°)
= 4 * (1/2) / (√3/2)
= 2 / √3
= (√3) / 3
Therefore, the ratio d/h is (√3) / 3.
To solve this problem, we need to use the equations of projectile motion. Let's break down the motion of the projectile into horizontal and vertical components.
The initial velocity of the projectile can be divided into horizontal and vertical components using trigonometry. Let's assume the initial velocity, magnitude V0, can be split into V0x (horizontal component) and V0y (vertical component).
V0x = V0 * cosθ
V0y = V0 * sinθ
Now, let's find the time it takes for the projectile to reach its maximum height. The vertical motion can be analyzed independently, considering only the vertical component (V0y) and the acceleration due to gravity (g).
Using the equation:
Vf = Vi + at
where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and t is time, we can find the time taken to reach maximum height.
At maximum height, the final vertical velocity (Vf) is 0, so the equation becomes:
0 = V0y − gt
Rearranging the equation, we get:
t = V0y / g
Now, let's determine the time it takes for the projectile to reach the ground after reaching its maximum height.
The time taken for the whole motion is equal to twice the time taken to reach the maximum height. This is because the time taken to reach the maximum height is the same as the time taken to fall from the maximum height to the ground.
So, the total time of flight (T) can be given as:
T = 2t = 2(V0y / g)
Now, let's find the horizontal distance (d) traveled by the projectile.
The horizontal distance traveled by the projectile can be determined using the equation:
d = V0x * T
Substituting the values from earlier equations, we get:
d = (V0 * cosθ) * (2(V0 * sinθ) / g)
Canceling out the common terms and simplifying, we have:
d = (V0^2 * sin(2θ)) / g
Finally, let's find the maximum height (h) reached by the projectile.
The maximum height can be determined using the equation:
h = (V0y^2) / (2g)
Substituting the values from earlier equations, we get:
h = (V0^2 * sin^2(θ)) / (2g)
Now, we can calculate the ratio d/h:
d/h = ((V0^2 * sin(2θ)) / g) / ((V0^2 * sin^2(θ)) / (2g))
Canceling out the common terms, we get:
d/h = 2 * sin(θ) * cos(θ) / sin^2(θ)
Using the double-angle identity, sin(2θ) = 2sin(θ)cos(θ), we simplify further to:
d/h = sin(2θ) / sin^2(θ)
Therefore, the ratio d/h is equal to sin(2θ) / sin^2(θ).
d =vₒ²•sin2α/g = 2•vₒ²•sinα•cosα/g,
h= vₒ²• sin² α/2g.
d/h =4/g•tan α = 4/9.8•tan60º=0.24.