To solve this problem, we need to break it into two parts: finding how long the ball was in the air (time) and determining the height from which it was thrown (initial vertical height).

Part 1: Time in the air

Since the ball is thrown horizontally, the only motion that affects the time in the air is the vertical motion under the influence of gravity. The horizontal velocity does not affect how long the ball stays in the air.

Using the formula for horizontal distance: $\text{Distance} = \text{Velocity} \times \text{Time}$ Given:

• Horizontal distance $d = 50 \, \text{m}$
• Horizontal velocity $v_x = 22 \, \text{m/s}$

We can rearrange the formula to solve for time: $t = \frac{d}{v_x} = \frac{50}{22} \approx 2.27 \, \text{seconds}$

Thus, the ball was in the air for approximately 2.27 seconds.

Part 2: Height of the throw

Now, to find the height from which the ball was thrown, we use the vertical motion equation: $y = \frac{1}{2} g t^2$ where:

• $y$ is the height (unknown),
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $t = 2.27 \, \text{seconds}$.

Substituting the values: $y = \frac{1}{2} \times 9.8 \times (2.27)^2$ $y \approx \frac{1}{2} \times 9.8 \times 5.15$ $y \approx 25.2 \, \text{m}$

So, the ball was thrown from a height of approximately 25.2 meters.

Summary:

• Time in the air: 2.27 seconds
• Height of throw: 25.2 meters

Would you like more details on any part of this?

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Here are some related questions for further exploration:

1. How would the result change if the ball was thrown at a different horizontal speed?
2. What would be the total velocity of the ball upon hitting the ground?
3. How does air resistance affect the motion of the ball in real-life situations?
4. What if the ball was thrown upwards at an angle instead of horizontally?
5. Can we calculate the range if the ball was thrown upwards?

Tip: Horizontal and vertical motions in projectile problems are independent of each other, except for the time they share.