Two spheres are launched horizontally from a 0.8 m high table. Sphere A is launched with an initial speed of 16 m/s. Sphere B is launched with an initial speed of 8 m/s.
To solve this problem, we need to use the equations of motion for projectile motion. Projectile motion is a type of motion in which an object is launched in the air and moves along a curved path under the influence of gravity.
The key equations for projectile motion are:
1. Horizontal motion equation:
distance = velocity × time
2. Vertical motion equations:
displacement = initial velocity × time + (1/2) × acceleration × time^2
final velocity^2 = initial velocity^2 + 2 × acceleration × displacement
Let's calculate some values for the two spheres:
For Sphere A:
- Initial vertical velocity (Vy) is zero because it is launched horizontally.
- Vertical displacement (Sy) is the height from which the sphere is launched, which is 0.8 m.
- Acceleration due to gravity (g) is approximately 9.8 m/s^2.
Using the equation for vertical displacement, we can calculate the time it takes for Sphere A to reach the ground:
0.8 = 0 × t + (1/2) × 9.8 × t^2
0.8 = 4.9t^2
t^2 = 0.8/4.9
t^2 = 0.1633
t ≈ √0.1633
t ≈ 0.404 s
Now that we have the time of flight, we can calculate the horizontal distance travelled by Sphere A using the equation for horizontal motion:
Distance_A = velocity × time
Distance_A = 16 m/s × 0.404 s
Distance_A ≈ 6.464 m
For Sphere B:
Sphere B is launched with half the initial speed of Sphere A, which means its horizontal speed is 8 m/s.
Since both spheres are launched horizontally, they have the same time of flight.
Using the equation for horizontal motion, we can calculate the horizontal distance travelled by Sphere B:
Distance_B = velocity × time
Distance_B = 8 m/s × 0.404 s
Distance_B ≈ 3.232 m
Therefore, Sphere A travels approximately 6.464 meters horizontally, while Sphere B travels approximately 3.232 meters horizontally.