A bean jumps off a table and when it reaches have its height it has a speed of 10 m/s, how high does it rise?
5.1m = 1/2 of h max. hmax = 10.2 m.
V^2 = Vo^2 + 2g*h.
10^2 = 0 + 19.6h, 19.6h = 100, h = 5.1 m.
To determine how high the bean rises, we can use the laws of projectile motion. The initial vertical velocity of the bean is zero since it starts from rest. The acceleration due to gravity, which acts downwards, is approximately 9.8 m/s².
We can use the equation:
vf² = vi² + 2ad
Where:
- vf is the final velocity (10 m/s)
- vi is the initial velocity (0 m/s)
- a is the acceleration due to gravity (-9.8 m/s²)
- d is the displacement (height)
Plugging in the known values:
10² = 0² + 2(-9.8)d
Simplifying:
100 = -19.6d
To solve for d, we can rearrange the equation:
d = 100 / -19.6
Calculating:
d ≈ -5.1 meters
The negative sign indicates that the displacement is upward, opposite to the direction of gravity. So, the bean rises approximately 5.1 meters before reaching half its maximum height.