If you throw a 0.5 kg ball straight up with a speed of 10 m/s, how high will it go?
v = Vi - 9.81 t
v = 0 at top
0 = 10 - 9.81 t
t = 1.02 seconds upward
h = Hi + Vi t - 4.9 t^2
h = 0 + 10(1.02) - 4.9(1.02)^2
h = 10.2 - 5.1 = 5.1 meters
note - this has nothing to do with the mass of the ball.
What exactly is the formula for it n jow did u get it?
To determine how high the ball will go when thrown straight up, you can use the principles of motion and the laws of mechanics. In this case, you can apply the equations of motion for an object in free fall.
Step 1: Identify the known values:
- Initial velocity (u) = 10 m/s (as it is thrown straight up)
- Final velocity (v) = 0 m/s (at the highest point)
- Acceleration (a) = -9.8 m/s² (due to gravity)
- Displacement (s) = ? (height reached)
Step 2: Determine the equation to use:
The equation to find displacement in the y-direction (height) can be derived from the second equation of motion:
v² = u² + 2as
Since the final velocity (v) is zero at the highest point, we can rewrite the equation as:
0 = u² + 2as
Step 3: Substitute the known values into the equation:
0 = (10 m/s)² + 2(-9.8 m/s²)s
Step 4: Simplify and solve for s (height):
0 = 100 m²/s² - 19.6 m/s²s
Rearranging the equation:
19.6 m/s²s = 100 m²/s²
Dividing both sides by 19.6 m/s²:
s = 100 m²/s² / 19.6 m/s² ≈ 5.102 m
Therefore, the ball will reach a height of approximately 5.102 meters when thrown straight up with a speed of 10 m/s.