A ball is kicked with an initial velocity of 25 s at an angle of 35° to the ground. Calculate: (a) The maximum height reached. (b) The time of flight. (c) The range.

The horizontal component of velocity ($v_x$) remains constant throughout the projectile's flight, assuming no air resistance.
$$
v_x = v_0 \cos \theta
$$

$$
v_x = 20 \text{ m/s} \times \cos 30^\circ
$$
$$
v_x = 20 \text{ m/s} \times \frac{\sqrt{3}}{2}
$$
$$
v_x \approx 17.32 \text{ m/s}
$$
The horizontal component of the velocity at maximum height is $\boxed{\text{17.32 m/s}}$.

At the maximum height of a projectile's trajectory, the vertical component of its velocity momentarily becomes zero.  
The vertical component of the velocity at maximum height is $\boxed{\text{0 m/s}}$.

Given: Maximum horizontal distance (Range), $R = 50 \text{ m}$.  
Initial velocity, $v_0 = 20 \text{ m/s}$.  
Launch angle, $\theta = 30^\circ$.  
The horizontal component of velocity is constant: $v_x = v_0 \cos \theta$.  
The time of flight ($T$) is related to the range by $R = v_x T$.

$$
v_x = v_0 \cos \theta = 20 \text{ m/s} \times \cos 30^\circ
$$
$$
v_x = 20 \text{ m/s} \times 0.866
$$
$$
v_x = 17.32 \text{ m/s}
$$

$$
T = \frac{R}{v_x}
$$
$$
T = \frac{50 \text{ m}}{17.32 \text{ m/s}}
$$
$$
T \approx 2.89 \text{ s}
$$
The time of flight is $\boxed{\text{2.89 s}}$.

The vertical displacement is $h = 1.5 \text{ m}$. The initial vertical velocity $v_{0y} = 0 \text{ m/s}$.
$$
h = v_{0y}t + \frac{1}{2}gt^2
$$

$$
1.5 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2
$$
$$
1.5 = 4.9t^2
$$

$$
t^2 = \frac{1.5}{4.9}
$$
$$
t = \sqrt{\frac{1.5}{4.9}}
$$
$$
t \approx 0.553 \text{ s}
$$
The time taken to reach the floor is $\boxed{\text{0.55 s}}$.

The horizontal distance ($R$) is given by $R = v_{0x}t$.

Using the time $t \approx 0.553 \text{ s}$ from part (a) and $v_{0x} = 2 \text{ m/s}$.
$$
R = 2 \text{ m/s} \times 0.553 \text{ s}
$$
$$
R \approx 1.11 \text{ m}
$$
The ball lands approximately $\boxed{\text{1.11 m}}$ from the table.

The horizontal velocity ($v_x$) remains constant:
$$
v_x = v_{0x} = 2 \text{ m/s}
$$
The vertical velocity ($v_y$) at impact is:
$$
v_y = v_{0y} + gt
$$
$$
v_y = 0 \text{ m/s} + (9.8 \text{ m/s}^2)(0.553 \text{ s})
$$
$$
v_y \approx 5.42 \text{ m/s}
$$

The impact velocity ($v$) is the resultant of its horizontal and vertical components:
$$
v = \sqrt{v_x^2 + v_y^2}
$$
$$
v = \sqrt{(2 \text{ m/s})^2 + (5.42 \text{ m/s})^2}
$$
$$
v = \sqrt{4 + 29.38}
$$
$$
v = \sqrt{33.38}
$$
$$
v \approx 5.78 \text{ m/s}
$$
The impact velocity of the ball is approximately $\boxed{\text{5.78 m/s}}$.

$$
H = \frac{v_0^2 \sin^2 \theta}{2g}
$$

Given $v_0 = 25 \text{ m/s}$, $\theta = 35^\circ$, $g = 9.8 \text{ m/s}^2$.
$$
\sin 35^\circ \approx 0.5736
$$
$$
H = \frac{(25 \text{ m/s})^2 (0.5736)^2}{2 \times 9.8 \text{ m/s}^2}
$$
$$
H = \frac{625 \times 0.3289}{19.6}
$$
$$
H = \frac{205.5625}{19.6}
$$
$$
H \approx 10.49 \text{ m}
$$
The maximum height reached is $\boxed{\text{10.49 m}}$.

$$
T = \frac{2 v_0 \sin \theta}{g}
$$

$$
\sin 35^\circ \approx 0.5736
$$
$$
T = \frac{2 \times 25 \text{ m/s} \times 0.5736}{9.8 \text{ m/s}^2}
$$
$$
T = \frac{50 \times 0.5736}{9.8}
$$
$$
T = \frac{28.68}{9.8}
$$
$$
T \approx 2.93 \text{ s}
$$
The time of flight is $\boxed{\text{2.93 s}}$.

$$
R = \frac{v_0^2 \sin(2\theta)}{g}
$$

$$
2\theta = 2 \times 35^\circ = 70^\circ
$$
$$
\sin 70^\circ \approx 0.9397
$$
$$
R = \frac{(25 \text{ m/s})^2 \sin(70^\circ)}{9.8 \text{ m/s}^2}
$$
$$
R = \frac{625 \times 0.9397}{9.8}
$$
$$
R = \frac{587.3125}{9.8}
$$
$$
R \approx 59.93 \text{ m}
$$
The range is $\boxed{\text{59.93 m}}$.