A ball is shot straight up into the air with initial velocity of 45 ft/sec. Assuming that the air resistance can be ignored, how high does it go?
Well, the ball could go really high, but it would be counterproductive for it to reach space. We wouldn't want any interstellar basketball games, would we? Anyway, let's get back to your question. To find out how high the ball goes, we can use the good old kinematic equations. We know the initial velocity is 45 ft/sec, but we don't know the time it takes for the ball to reach its maximum height. So, I'll just sit here and entertain you while you do the math. Good luck!
To determine how high the ball goes, we can use the kinematic equation for the displacement of an object in free fall:
s = ut - (1/2)gt^2
where:
- s is the displacement (in this case, the height)
- u is the initial velocity
- t is the time
- g is the acceleration due to gravity (approximately 32 ft/sec²)
Since the ball is shot straight up, its initial velocity is positive (+45 ft/sec). We need to find the time it takes for the ball to reach its highest point. At the highest point, the final velocity will be 0.
Using the equation for the final velocity in free fall:
v = u - gt
where:
- v is the final velocity
- t is the time
We can rearrange the equation to solve for the time:
t = u / g
Substituting the given values:
t = 45 ft/sec / 32 ft/sec² ≈ 1.40625 sec
Now, we can find the maximum height reached by substituting this time value into the displacement equation:
s = ut - (1/2)gt^2
s = (45 ft/sec)(1.40625 sec) - (1/2)(32 ft/sec²)(1.40625 sec)²
s ≈ 31.6406 ft
Therefore, the ball reaches a height of approximately 31.6406 feet.
To find out how high the ball goes, we need to calculate the maximum height it reaches. We can solve this problem using the laws of motion and the kinematic equations.
The first step is to determine the time it takes for the ball to reach its maximum height. We can use the fact that the vertical velocity of the ball decreases due to the acceleration of gravity. At the highest point, the ball's velocity becomes zero.
The equation that relates initial velocity, final velocity, acceleration, and time is:
v = u + at
where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, the initial velocity (u) is 45 ft/sec, and the final velocity (v) is 0 ft/sec at the top of its trajectory.
So, we have:
0 = 45 - 32.2t
Solving this equation for t, we get:
t = 45 / 32.2
t ≈ 1.398 seconds
This tells us that it takes approximately 1.398 seconds for the ball to reach its maximum height.
Now, we can use this time to find the maximum height that the ball reaches. We can use the equation:
s = ut + (1/2)at^2
where:
- s is the displacement (maximum height),
- u is the initial velocity,
- a is the acceleration (always equal to the acceleration due to gravity), and
- t is the time.
Plugging in the values, we get:
s = 45 * 1.398 + (1/2) * 32.2 * (1.398)^2
Simplifying this equation, we find:
s ≈ 87.41 ft
Therefore, the ball reaches a maximum height of approximately 87.41 feet.