A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 15.0 m/s, at an angle of theta = 53.0 degrees from the horizontal. The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down. Calculate how long the ball is in the air. ( g = 9.81 m/s2)

Vo = (15m/ws,53deg.).

Xo = 15cos53 = 9.0mm/s.
Yo = 15sin53 = 12.0m/s.

t(up) = (Yf - Yo) / g,
t(up) = (0 - 12) / -9.81) = 1.22s.

Yf^2 = Yo^2 + 2gh,
h = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (12)^2) / -19.62 = 7.34m.

h = Yo*t + 0.5g*t^2 = (7.34 - 2)m.
0 + 4.9t^2 = 5.34,
t^2 = 1.09,
t(dn) = 1.04s.

T = t(up) + t(dn) = 1.22 + 1.04 = 2.26s
= Time in air.

To solve this problem, we can use the equations of motion for projectile motion. Let's break down the problem into two components: horizontal and vertical.

First, let's find the time it takes for the ball to reach its highest point. Since the initial vertical velocity is given by the initial speed multiplied by the sine of the launch angle, we can calculate it as follows:

Vertical component of initial velocity = v₀y = (15.0 m/s) * sin(53.0°)

Using this vertical component of initial velocity, we can now find the time it takes for the ball to reach its highest point using the equation:

Time to reach highest point = tᵤ = v₀y / g

where g is the acceleration due to gravity (9.81 m/s²).

Next, let's find the time it takes for the ball to fall back to the height of the father's hand (2.0 m). To do this, we can use the equation:

Vertical distance = v₀yt - 0.5 * g * t²

where v₀yt is the vertical component of the initial velocity multiplied by the time to reach the highest point. Solving for time, we get:

tᵠ = √((2 * v₀yt) / g)

Finally, the total time the ball is in the air can be found by adding the time it takes to reach the highest point (tᵤ) and the time it takes to fall back to the height of the father's hand (tᵠ):

Total time in the air = t = tᵤ + tᵠ

Substituting the known values into these equations will give us the answer to the problem.