Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 22 m/s at an angle 63 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

To determine the time it takes for the ball to reach Sarah, we can use the kinematic equations. The key equation we will use is the vertical motion equation:

y = y0 + V0y * t - (1/2) * g * t^2

where:
y = final vertical position (height of Sarah's hand) = 1.5 meters
y0 = initial vertical position (height at which Julie released the ball) = 1.5 meters
V0y = vertical component of the initial velocity = V0 * sin(theta), where V0 is the initial speed (22 m/s) and theta is the angle (63 degrees)
g = acceleration due to gravity = 9.8 m/s^2 (assuming no air resistance)
t = time

Substituting the values into the equation, we have:

1.5 = 1.5 + (22 * sin(63)) * t - (1/2) * (9.8) * t^2

Simplifying this equation gives us a quadratic equation:

0 = (-4.9) * t^2 + (22 * sin(63)) * t

To solve for t, we can set this equation equal to zero and factor it:

0 = t * [(-4.9) * t + (22 * sin(63))]

Using the zero product property, we can set each factor equal to zero:

t = 0 or (-4.9) * t + (22 * sin(63)) = 0

The first solution, t = 0, can be disregarded since it represents the initial moment when Julie throws the ball. Solving the second equation gives us:

(-4.9) * t + (22 * sin(63)) = 0

t = (22 * sin(63)) / 4.9

Calculating this value gives us the time it takes for the ball to reach Sarah.

To find the time it takes for the ball to reach Sarah, we can use the vertical motion equation:

y = y0 + v0y * t - 0.5 * g * t^2

Since the initial vertical velocity is given by v0y = v0 * sin(θ), where θ is the angle with respect to the horizontal, and the initial vertical position is y0 = 1.5 m, we can rewrite the equation as:

y = 1.5 + (v0 * sin(θ)) * t - 0.5 * g * t^2

Since the ball arrives at the same height it was thrown from, y = 1.5 m. And since the acceleration due to gravity is g = 9.8 m/s^2, we have:

1.5 = 1.5 + (22 * sin(63)) * t - 0.5 * (9.8) * t^2

Simplifying the equation:

0 = 13.85 * t - 4.9 * t^2

Now we can solve for t by using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Where a = -4.9, b = 13.85, and c = 0. Solving for t:

t = (-13.85 ± sqrt(13.85^2 - 4 * -4.9 * 0)) / (2 * -4.9)

t = (-13.85 ± sqrt(192.4225)) / (-9.8)

Using the positive value of t:

t = (-13.85 + sqrt(192.4225)) / (-9.8)

t ≈ 2.12 seconds

Therefore, it takes approximately 2.12 seconds for the ball to reach Sarah.

horizontal speed = u = 22 cos 63 the whole time

Vi = 22 sin 63

at top v = 0 = Vi - g t where t is rise time t = Vi/9.81

average speed up = Vi/2
so top height = (Vi/2)t + 1.5

total time in air = 2 t
so Sarah is u (2t) = 22 coa 63 (2t) meters from Julie