A metal sphere is launched with an initial velocity of 1.5 m/s as it leaves the ramp. The end of the ramp is 1.2 m above the floor. Calculate the range of the sphere.
launched horizontally?
time to fall 1.2 m
1.2=1/2 g t^2
t=sqrt(2.4/g)
distance hoizontal=1.5*time t
To calculate the range of the sphere, we need to find the horizontal distance traveled by the sphere.
The vertical motion of the sphere can be analyzed using the equation:
y = yo + voy * t + (1/2) * a * t^2
where:
- y is the final vertical position (0 m, as the sphere hits the floor)
- yo is the initial vertical position (1.2 m, as the sphere is launched from the ramp)
- voy is the initial vertical velocity (0 m/s, as the sphere is launched horizontally)
- a is the acceleration due to gravity (-9.8 m/s^2, as the sphere falls vertically downward)
- t is the time taken for the sphere to hit the floor
Using the equation, we can solve for t:
0 = 1.2 + 0 * t + (1/2) * (-9.8) * t^2
Simplifying the equation:
4.9 * t^2 = 1.2
t^2 = 1.2 / 4.9
t^2 ≈ 0.2449
Taking the square root of both sides:
t ≈ √0.2449
t ≈ 0.4949 seconds (approximately)
Now, let's calculate the horizontal distance traveled by the sphere using the equation:
x = vox * t
where:
- x is the horizontal distance traveled (range)
- vox is the initial horizontal velocity (1.5 m/s, as given in the question)
- t is the time we calculated earlier (0.4949 seconds)
Substituting the values into the equation:
x = 1.5 * 0.4949
x ≈ 0.74235 meters (approximately)
Therefore, the range of the sphere is approximately 0.74235 meters.
To calculate the range of the sphere, we need to find the horizontal distance traveled by the sphere before hitting the floor. We can use the equations of motion to solve this problem.
First, let's determine the time it takes for the sphere to reach the ground. We need to find the time it takes for the sphere to fall from the end of the ramp to the floor.
Using the equation for the vertical displacement of an object in free fall:
h = vit + (1/2)gt^2
Where:
h = vertical displacement
vi = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the sphere is launched horizontally, the initial vertical velocity (vi) is 0 m/s, and the vertical displacement (h) is 1.2 m. Plugging in these values, the equation becomes:
1.2 = 0 + (1/2)(9.8)(t^2)
1.2 = 4.9t^2
Simplifying the equation gives us:
t^2 = 1.2 / 4.9
t^2 = 0.2449
Taking the square root of both sides, we find:
t ≈ 0.4949 seconds
Now that we know the time it takes for the sphere to hit the floor, we can calculate the horizontal distance traveled by the sphere using the equation:
d = vxt
Where:
d = horizontal distance (range)
vx = horizontal velocity
t = time
Given that the initial horizontal velocity (vx) is 1.5 m/s and the time (t) is 0.4949 seconds, we can substitute these values into the equation:
d = 1.5 m/s * 0.4949 s
d ≈ 0.7424 meters
Therefore, the range of the metal sphere is approximately 0.7424 meters.