A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 2.80 s for the ball to reach its maximum height.
(a) Find its initial velocity. m/s (upward)
(b) Find the height it reaches.
The correct answer and explanation is :
We will solve this using kinematic equations. Given that the baseball takes 2.80 s to reach its maximum height, we analyze the motion under gravity.
Step 1: Understanding the Motion
- When the ball is at its maximum height, its velocity becomes zero (( v = 0 )).
- The acceleration due to gravity acts downward and has a magnitude of ( g = 9.81 \, \text{m/s}^2 ).
- The time taken to reach the peak is ( t = 2.80 ) s.
- Using the kinematic equation: [
v = u + at
] where: - ( v ) is the final velocity (0 m/s at max height),
- ( u ) is the initial velocity (to be found),
- ( a = -9.81 ) m/s² (negative because gravity acts downward),
- ( t = 2.80 ) s.
(a) Finding Initial Velocity
Using the equation:
[
0 = u – (9.81)(2.80)
]
[
u = (9.81)(2.80)
]
[
u = 27.47 \, \text{m/s}
]
So, the initial velocity is 27.47 m/s (upward).
(b) Finding Maximum Height
Using another kinematic equation:
[
v^2 = u^2 + 2as
]
Since ( v = 0 ) at max height,
[
0 = u^2 + 2(-9.81) s
]
[
s = \frac{u^2}{2g} = \frac{(27.47)^2}{2(9.81)}
]
[
s = \frac{754.56}{19.62}
]
[
s = 38.47 \, \text{m}
]
So, the maximum height reached is 38.47 m.
Explanation (300 words)
The problem involves one-dimensional motion under gravity. When the baseball is hit, it moves upward until it reaches a maximum height, where its velocity becomes zero before coming back down. The time taken to reach the highest point is given as 2.80 seconds. Using this information, we can calculate its initial velocity by applying the first kinematic equation:
[
v = u + at
]
Since the velocity at the peak is zero, we rearrange to solve for ( u ), considering gravity as ( -9.81 \, \text{m/s}^2 ). Plugging in the values, we find that the initial velocity is 27.47 m/s.
To find the maximum height, we use another kinematic equation that relates velocity, acceleration, and displacement. Since the final velocity at the peak is zero, we use:
[
v^2 = u^2 + 2as
]
Solving for ( s ), we substitute the values and get 38.47 meters.
Thus, the baseball reaches a height of 38.47 m before coming back down.