Question

A ball thrown straight upward returns to its original level in 2.5 seconds. A second ball is thrown at an angle of 40 degrees above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.5 seconds?

Answer 1

The initial speed of the second ball, thrown at an angle of 40 degrees above the horizontal and returning to its original level in 2.5 seconds, is approximately 18.74 m/s.

1. For the first ball (thrown straight upward),
`\( v_(01) = gt/2 \)`, where t = 2.5 seconds:

`\[ v_(01) = (9.8 \ \text{m/s}^2) * (2.5 \ \text{s}) / 2 = 12.25 \ \text{m/s} \]`

2. Now, use the relationship
`\( v_(01) = v_(02) \sin(\theta) \)` to find
`\( v_(02) \)` for the second ball (thrown at 40 degrees):

`\[ v_(02) = (v_(01))/(\sin(\theta)) = \frac{12.25 \ \text{m/s}}{\sin(40^\circ)} \]`

The final result for
`\( v_(02) \)` is approximately 18.74 m/s (rounded to two decimal places).

Both balls, one thrown straight upward and the other at a 40-degree angle, return to the original level in 2.5 seconds. The initial speed of the angled throw is approximately 18.74 m/s, ensuring matching time of flight.