A ball thrown straight upward returns to its original level in 2.30 . A second ball is thrown at an angle of 38.0 degrees above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.30 ?

To find the initial speed of the second ball, we can use the kinematic equations of motion.

Let's first calculate the time it takes for the second ball to reach its maximum height. Since it returns to its original level in the same time as the first ball (2.30 s), the time taken for the second ball to reach its maximum height (on the way up) will be half of that, which is 1.15 s.

Using the equation for vertical displacement during constant acceleration:

y = v₀y * t - 0.5 * g * t²

Where:
y = vertical displacement
v₀y = initial vertical velocity = v₀ * sin(θ)
t = time
g = acceleration due to gravity = 9.8 m/s²

At the maximum height, the vertical displacement is 0 since the ball has returned to its original level. Therefore:

0 = (v₀ * sin(θ)) * (1.15) - 0.5 * (9.8) * (1.15)²

Simplifying the equation, we have:

0 = 1.15 * v₀ * sin(θ) - 6.66 * 1.15²

Next, let's calculate the time taken for the second ball to reach the ground after reaching its maximum height. Again, this time will be the same as the first ball (2.30 s).

Using the equation for vertical displacement during constant acceleration:

0 = (v₀ * sin(θ)) * (2.30) - 0.5 * (9.8) * (2.30)²

Simplifying the equation, we have:

0 = 2.30 * v₀ * sin(θ) - 6.66 * 2.30²

Now, since the second ball returns to its original level, we can equate the two equations:

1.15 * v₀ * sin(θ) - 6.66 * 1.15² = 2.30 * v₀ * sin(θ) - 6.66 * 2.30²

Solving for v₀, we get:

v₀ = (6.66 * 2.30²) / (2.30 * sin(θ) - 1.15 * sin(θ))

Now we can substitute the given values: θ = 38.0 degrees.

v₀ = (6.66 * 2.30²) / (2.30 * sin(38.0) - 1.15 * sin(38.0))

Calculating the expression on the right side of the equation will give us the initial speed of the second ball.