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A ball is thrown vertically upward with a speed of 25.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started?

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Answer:

Step-by-step explanation:

Let's address each part of the problem separately:

(a) To find out how high the ball rises, we can use the following kinematic equation for vertical motion:

\[h = \frac{v^2}{2g}\]

where:

- \(h\) is the height,

- \(v\) is the initial vertical velocity (25.0 m/s),

- \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).

Now, plug in the values:

\[h = \frac{(25.0\, \text{m/s})^2}{2(9.81\, \text{m/s}^2)}\]

Calculate \(h\):

\[h \approx 31.88\, \text{meters}\]

So, the ball rises to a height of approximately 31.88 meters.

(b) To find the time it takes to reach its highest point, you can use the following equation:

\[t = \frac{v}{g}\]

where:

- \(t\) is the time,

- \(v\) is the initial vertical velocity (25.0 m/s),

- \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).

Now, plug in the values:

\[t = \frac{25.0\, \text{m/s}}{9.81\, \text{m/s}^2}\]

Calculate \(t\):

\[t \approx 2.55\, \text{seconds}\]

So, it takes approximately 2.55 seconds to reach the highest point.

(c) To find the time it takes for the ball to hit the ground after reaching its highest point, we can use the fact that the time it takes to reach the highest point is the same as the time it takes to fall back down from that point. So, the total time of flight is twice the time calculated in part (b):

Total time = \(2 \times 2.55\, \text{seconds} = 5.10\, \text{seconds}\)

(d) To find the velocity when the ball returns to the level from which it started, we can use the following equation:

\[v = v_0 - gt\]

where:

- \(v\) is the final velocity (the velocity when it returns to the starting point),

- \(v_0\) is the initial velocity (25.0 m/s),

- \(g\) is the acceleration due to gravity (approximately 9.81 m/s²),

- \(t\) is the total time of flight (5.10 seconds).

Now, plug in the values:

\[v = 25.0\, \text{m/s} - (9.81\, \text{m/s}^2)(5.10\, \text{seconds})\]

Calculate \(v\):

\[v \approx -49.5\, \text{m/s}\]

The negative sign indicates that the velocity is in the opposite direction of the initial velocity. So, the velocity when the ball returns to the level from which it started is approximately 49.5 m/s downward.

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