A ball is thrown vertically upward with a
speed of 27.6 m/s.
How high does it rise?
Answer in units of m.
Vi^2=2g*h solve for h.
78
To determine the height the ball rises, we can use the equations of motion. The ball is thrown vertically, so we only need to consider the vertical motion.
We can start by using the equation: vf = vi + at, where
- vf is the final velocity (0 m/s at the highest point since the ball momentarily comes to rest),
- vi is the initial velocity (27.6 m/s),
- a is the acceleration (due to gravity, which is approximately -9.8 m/s²).
Plugging in the values, we can solve for t (the time it takes for the ball to reach its highest point):
0 = 27.6 m/s + (-9.8 m/s²) * t
Rearranging the equation:
9.8 m/s² * t = 27.6 m/s
Solving for t:
t = 27.6 m/s / 9.8 m/s²
t ≈ 2.82 s
Next, we can determine the height the ball reaches using the equation:
h = vi * t + (1/2) * a * t², where
- h is the height,
- vi is the initial velocity (27.6 m/s),
- t is the time (2.82 s),
- a is the acceleration (approximately -9.8 m/s²).
Plugging in the values, we can solve for h:
h = (27.6 m/s) * (2.82 s) + (1/2) * (-9.8 m/s²) * (2.82 s)²
Simplifying the equation:
h ≈ 78.1 m
Therefore, the ball rises to a height of approximately 78.1 meters.