Final answer:
To find the horizontal velocity of the box, we calculated the time it took to fall to the ground and then used this time to calculate the horizontal velocity. The calculated horizontal velocity was found to be approximately 1.07 m/s.
Step-by-step explanation:
The question asks for the horizontal velocity of the box at the moment it left the table top. To find this, we need to apply the kinematic equations for projectile motion. The box falls under gravity and lands 0.452 m away from the table, with a height of 0.875 m.
First, we calculate the time 't' it takes for the box to fall to the floor using the vertical motion formula:
s = ut + ½at²
Where:
- s is the vertical displacement (0.875 m)
- u is the initial vertical velocity (0)
- a is the acceleration due to gravity (9.81 m/s²)
0.875 m = 0 + 0.5 * 9.81 m/s² * t²
Solving for 't' gives us the time it takes for the box to fall:
t = sqrt((2 * 0.875 m) / 9.81 m/s²) = 0.423 s
Using this time, we can find the horizontal velocity 'v' using the horizontal displacement formula:
s = vt
Where:
- s is the horizontal displacement (0.452 m)
- v is the horizontal velocity we need to find
- t is the time (0.423 s)
0.452 m = v * 0.423 s
Solving for 'v' gives us the horizontal velocity:
v = 0.452 m / 0.423 s = 1.07 m/s