A projectile starts from rest and moves 7 m down a frictionless ramp inclined at 21◦ with the horizontal. The acceleration of gravity is 9.8 m/s2 . What will be the range of the projectile if the bottom of the ramp is 1.7 m above the ground? Answer in units of m.
To find the range of the projectile, we need to analyze its motion along the horizontal direction.
The first step is to determine the time it takes for the projectile to reach the bottom of the ramp. We can use the kinematic equation for motion in a straight line:
Δy = v₀y * t + (1/2) * a * t²
Since the projectile starts from rest and moves vertically downward, its initial vertical velocity (v₀y) will be 0. The vertical acceleration (a) is equal to the acceleration due to gravity (-9.8 m/s²) because the ramp is frictionless.
Substituting the given values into the equation, we get:
1.7 m = 0 * t + (1/2) * (-9.8 m/s²) * t²
Rearranging the equation, we have:
4.9t² = 1.7 m
Dividing both sides by 4.9, we find:
t² = 0.3469 s²
Taking the square root of both sides, we obtain:
t ≈ 0.589 s
So, it takes approximately 0.589 seconds for the projectile to reach the bottom of the ramp.
Next, we can find the horizontal distance traveled by the projectile during this time. The horizontal velocity of the projectile remains constant because there are no horizontal forces acting on it. We can use the formula:
Δx = v₀x * t
To find the initial horizontal velocity (v₀x), we can use trigonometry. We know that the ramp is inclined at an angle of 21° with the horizontal. The relationship between the horizontal and vertical components of velocity is given by:
v₀x = v₀ * cos(θ)
Where v₀ is the initial velocity of the projectile and θ is the angle of the ramp. Since the projectile starts from rest, its initial velocity will be 0.
v₀x = 0 * cos(21°) = 0 m/s
Substituting this value into the formula, we find:
Δx = 0 * t = 0 m
Therefore, the range of the projectile is 0 meters.
To find the range of the projectile, we need to determine the horizontal distance traveled by the projectile when it reaches the ground. We can break down the problem into two parts: the motion on the inclined ramp and the free fall motion.
First, let's find the time taken for the projectile to reach the bottom of the ramp:
The vertical distance traveled is the height of the ramp, which is given as 1.7 m.
The vertical acceleration is due to gravity, which is 9.8 m/s^2.
Using the equation of motion for vertical motion:
Δy = V₀y * t + (1/2) * a * t²
Since the projectile starts from rest, V₀y = 0, and the equation becomes:
1.7 m = 0 * t + (1/2) * 9.8 m/s² * t²
Simplifying the equation:
1.7 m = 4.9 m/s² * t²
Dividing both sides by 4.9 m/s²:
0.3469 s² = t²
Taking the square root of both sides:
t = 0.59 s
Now, let's find the horizontal distance traveled:
The horizontal distance traveled is the range we need to find.
The horizontal acceleration is 0 m/s² because there is no resistance or friction.
The horizontal velocity is constant throughout the motion on the inclined ramp.
Using the equation of motion for horizontal motion:
Δx = V₀x * t + (1/2) * a * t²
Since the horizontal acceleration is 0 m/s², the equation becomes:
Δx = V₀x * t
To find the initial horizontal velocity (V₀x), we can use the trigonometric relationship between the components of the initial velocity and the inclination angle of the ramp:
V₀x = V₀ * cos(θ)
Where V₀ is the initial velocity and θ is the inclination angle (21°).
The initial velocity (V₀) can be found using the equation of motion for vertical motion:
V₀y = V₀ * sin(θ)
Since V₀y = 0 at the start of motion, we have:
0 = V₀ * sin(21°)
Solving for V₀:
V₀ = 0 / sin(21°)
V₀ = 0 m/s
Therefore, V₀x = V₀ * cos(θ)
V₀x = 0 * cos(21°)
V₀x = 0 m/s
Substituting V₀x = 0 m/s and t = 0.59 s into the equation:
Δx = 0 * 0.59 s
Therefore, the range of the projectile on the horizontal ground is 0 meters.