A ball is thrown straight upward and returns to the thrower's hand after 3.40 s in the air. A second ball is thrown at an angle of 29.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

To solve this problem, we can use the kinematic equations of motion. Let's break down the problem into two parts:

1. Vertical motion of the first ball:
The first ball is thrown vertically, so we can use the equation for the height of an object thrown upwards:
h = v0t - (1/2)gt^2

Here, h is the maximum height reached by the first ball, v0 is its initial velocity, t is the time it takes to reach the maximum height, and g is the acceleration due to gravity.

Given that the ball returns to the thrower's hand at a time of 3.40 s, we can substitute the values into the equation:
0 = v0(3.40) - (1/2)(9.8)(3.40)^2

Simplifying the equation will give us the value of v0, which is the initial velocity of the first ball.

2. Horizontal motion of the second ball:
The second ball is thrown at an angle of 29.0° with the horizontal. We know that the time it takes for the second ball to reach the same height as the first ball is also 3.40 s.

Now, we can use the horizontal component of the projectile's velocity to find its initial velocity. We know that the horizontal and vertical components of velocity are related by the equation:
v0x = v0 cos(θ)

Here, v0x is the horizontal component of the initial velocity, v0 is the magnitude of the initial velocity, and θ is the angle of projection.

Since we're given the angle of projection (29.0°) and the time of flight (3.40 s), we can use the equation:
distance = v0x * t

To cover the same vertical distance as the first ball, the distance traveled by the second ball in the vertical direction would be equal to the maximum height reached by the first ball.

Therefore,
h = v0x * t

Substituting the values of t and h, we can solve for v0x.

Once we have v0x, we can find the magnitude of the initial velocity (v0) by using the equation:
v0 = v0x / cos(θ)

Thus, we can calculate the required initial velocity (v0) of the second ball to reach the same height as the first ball.

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