a ball is kicked with initial velocity of 20 m/s in the horizontal drection and 16 m/s in the vertical direction.  assuming the ball was kicked from the ground.  At what speed (in m/s) does the ball hit the ground?

To determine the speed at which the ball hits the ground, we need to analyze the motion in both the horizontal and vertical directions. Since there is no air resistance, the horizontal and vertical motions are independent of each other.

### Horizontal Motion:
- The horizontal velocity (\(v_x\)) remains constant throughout the flight because there is no horizontal acceleration (assuming no air resistance).
- \(v_x = 20 \, \text{m/s}\)

### Vertical Motion:
- The vertical velocity (\(v_y\)) changes due to the acceleration caused by gravity (\(g = 9.8 \, \text{m/s}^2\)).
- The initial vertical velocity (\(v_{y0}\)) is \(16 \, \text{m/s}\).
- The ball accelerates downward under gravity, and its vertical velocity at any time \(t\) is given by:
  \[
  v_y = v_{y0} + gt
  \]
  However, we are interested in the final velocity just before the ball hits the ground.

### Step 1: Calculate the time of flight
The ball starts at ground level and will return to the ground, so the vertical displacement is zero. Using the vertical motion equation for displacement:
\[
\Delta y = v_{y0} t - \frac{1}{2} g t^2
\]
Since the ball starts and ends at the same height (\(\Delta y = 0\)):
\[
0 = 16t - \frac{1}{2} (9.8) t^2
\]
This simplifies to:
\[
t (16 - 4.9 t) = 0
\]
The solutions are \(t = 0\) (at the start) and:
\[
t = \frac{16}{4.9} \approx 3.27 \, \text{seconds}
\]
This is the total time of flight.

### Step 2: Calculate the final vertical velocity
At the moment the ball hits the ground, the vertical velocity can be calculated as:
\[
v_y = v_{y0} - g t
\]
Substitute \(v_{y0} = 16 \, \text{m/s}\), \(g = 9.8 \, \text{m/s}^2\), and \(t = 3.27 \, \text{seconds}\):
\[
v_y = 16 - (9.8)(3.27) \approx 16 - 32.05 \approx -16.05 \, \text{m/s}
\]
The negative sign indicates that the velocity is downward.

### Step 3: Calculate the resultant speed
The resultant speed when the ball hits the ground is the combination of the horizontal and vertical velocities. The horizontal velocity remains constant at \(20 \, \text{m/s}\), and the vertical velocity is \(-16.05 \, \text{m/s}\).

Using the Pythagorean theorem, the resultant speed \(v_{\text{final}}\) is:
\[
v_{\text{final}} = \sqrt{v_x^2 + v_y^2}
\]
\[
v_{\text{final}} = \sqrt{(20)^2 + (-16.05)^2}
\]
\[
v_{\text{final}} = \sqrt{400 + 257.6} \approx \sqrt{657.6} \approx 25.64 \, \text{m/s}
\]

### Final Answer:
The ball hits the ground at a speed of approximately **25.64 m/s**.

Would you like further details or explanations on any step?

### Related Questions:
1. How would the time of flight change if the initial vertical velocity were doubled?
2. What would be the effect of air resistance on the final speed?
3. If the ball were kicked from a height, how would you calculate the time of flight?
4. What is the horizontal range of the ball in this case?
5. How do the horizontal and vertical motions differ in projectile motion?

### Tip:
In projectile motion, the horizontal velocity remains constant, while the vertical velocity changes due to gravity. Always treat the two components separately.

A ball is kicked with initial velocity of 20 m/s in the horizontal direction and 16 m/s in the vertical direction. Assuming the ball was kicked from the ground, at what speed (in m/s) does the ball hit the ground?

Learn how to calculate the speed of a ball as it hits the ground after being kicked with an initial velocity of 20 m/s horizontally and 16 m/s vertically. This problem covers key concepts in projectile motion and kinematics. Suitable for Grades 10-12.

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