 A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0.  What is the maximum height h that it achieves, and what time th did that take, using g = 9.80 m/s?  Show the final formulas for the quantities, along with the numerical answers.  


We can solve this problem using the equations of motion for uniformly accelerated motion. The two quantities we need to find are the maximum height (\( h \)) and the time to reach that height (\( t_h \)).

### 1. Maximum Height (\( h \)):
At the maximum height, the final velocity (\( v_f \)) of the ball becomes zero. We can use the following kinematic equation:

\[
v_f^2 = v_i^2 + 2a y
\]

Where:
- \( v_f = 0 \, \text{m/s} \) (final velocity at maximum height),
- \( v_i = 4.00 \, \text{m/s} \) (initial velocity),
- \( a = -g = -9.80 \, \text{m/s}^2 \) (acceleration due to gravity, acting downward),
- \( y = h \) (the maximum height).

Substituting the known values:

\[
0 = (4.00)^2 + 2(-9.80) h
\]

\[
0 = 16.00 - 19.60 h
\]

\[
19.60 h = 16.00
\]

\[
h = \frac{16.00}{19.60} = 0.816 \, \text{m}
\]

So, the maximum height \( h \) is:

\[
h = 0.816 \, \text{m}
\]

### 2. Time to Reach Maximum Height (\( t_h \)):

To find the time taken to reach the maximum height, we can use the following kinematic equation:

\[
v_f = v_i + at
\]

Where:
- \( v_f = 0 \, \text{m/s} \),
- \( v_i = 4.00 \, \text{m/s} \),
- \( a = -9.80 \, \text{m/s}^2 \),
- \( t = t_h \) (time to reach the maximum height).

Substituting the known values:

\[
0 = 4.00 + (-9.80) t_h
\]

\[
9.80 t_h = 4.00
\]

\[
t_h = \frac{4.00}{9.80} = 0.408 \, \text{seconds}
\]

Thus, the time to reach the maximum height \( t_h \) is:

\[
t_h = 0.408 \, \text{seconds}
\]

### Final Formulas:
- Maximum height \( h \):  
\[
h = \frac{v_i^2}{2g}
\]
- Time to reach maximum height \( t_h \):  
\[
t_h = \frac{v_i}{g}
\]

### Numerical Answers:
- Maximum height \( h = 0.816 \, \text{m} \)
- Time to reach maximum height \( t_h = 0.408 \, \text{seconds} \)

Would you like a more detailed explanation, or do you have any other questions? Here are 5 related questions you might find interesting:
1. How would the maximum height change if the initial speed were doubled?
2. What happens if the ball is thrown downward instead of upward? 
3. How do air resistance and drag affect the motion of a projectile?
4. How can we derive the equations of motion for uniformly accelerated motion from calculus?
5. What is the total time for the ball to return to its original position after being thrown up?

**Tip:** In projectile motion problems with only gravity acting, the time to rise to the maximum height is always equal to the time to fall back to the starting point (ignoring air resistance).

A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and what time th did that take, using g = 9.80 m/s? Show the final formulas for the quantities, along with the numerical answers.

Explore how to calculate the maximum height and time to reach maximum height for a ball thrown upwards with an initial speed of 4.00 m/s. The solution uses key kinematic equations and concepts of projectile motion. Suitable for high school students learning kinematics.

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