Two spheres are launched horizontally from a 1.1 m -high table. Sphere A is launched with an initial speed of 4.5 m/s . Sphere B is launched with an initial speed of 1.5 m/s . What is the distance that sphere A travels from the edge of the table? What is the distance that sphere B travels from the edge of the table?
first, figure how long it takes to fall 1.1 meters.
s = 4.9t^2
Then, using that t, figure how far horizontally the travel, since
s = vt
Naturally, since B's speed is 1/3 of A's, it will travel 1/3 as far, since both balls take the same time to drop to the floor.
To find the horizontal distance traveled by the spheres from the edge of the table, we need to determine the time of flight for each sphere and multiply it by their respective horizontal velocities.
Let's start with Sphere A:
1. To find the time of flight, we need to determine how long it takes for Sphere A to hit the ground. Since the vertical motion of the sphere is subject to the force of gravity, we can use the formula:
h = v₀t + (1/2)gt²
where h is the vertical displacement (1.1 m), v₀ is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight.
Plugging in the values, we get:
1.1 m = 0 m/s ⋅ t + (1/2) ⋅ (-9.8 m/s²) ⋅ t²
Simplifying the equation:
1.1 m = -(4.9 m/s²) ⋅ t²
Rearranging the equation:
t² = -1.1 m / -(4.9 m/s²)
t² = 0.2245 s²
Taking the square root of both sides:
t ≈ 0.474 s
2. The horizontal distance traveled by Sphere A is given by the equation:
d = v₀x ⋅ t
where v₀x is the horizontal component of the initial velocity (4.5 m/s) and t is the time of flight (0.474 s).
Plugging in the values, we get:
d = 4.5 m/s ⋅ 0.474 s
d ≈ 2.133 m
Therefore, Sphere A travels approximately 2.133 meters from the edge of the table.
Now let's calculate the distance traveled by Sphere B using the same approach.
1. The time of flight can be found using the same formula as before, where h is the vertical displacement (1.1 m), v₀ is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s²), and t is the time of flight:
1.1 m = 0 m/s ⋅ t + (1/2) ⋅ (-9.8 m/s²) ⋅ t²
Simplifying the equation:
1.1 m = -(4.9 m/s²) ⋅ t²
Rearranging the equation:
t² = -1.1 m / -(4.9 m/s²)
t² = 0.2245 s²
Taking the square root of both sides:
t ≈ 0.474 s
2. The horizontal distance traveled by Sphere B can be calculated using the same equation:
d = v₀x ⋅ t
where v₀x is the horizontal component of the initial velocity (1.5 m/s) and t is the time of flight (0.474 s).
Plugging in the values, we get:
d = 1.5 m/s ⋅ 0.474 s
d ≈ 0.711 m
Therefore, Sphere B travels approximately 0.711 meters from the edge of the table.