An arrow is launched upward with an initial speed of 70m/s ( meters per second). The equation `v^(2)=v_(0)^(2)-2gh` describes the motion of the arrow, where `v_(0)` is the initial speed of the arrow, `v` is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity ( approximately `9.8 m//s^(2)`). What is the maximum height from the ground the arrow will rise to the nearest meter?

Answer

To find the maximum height the arrow will reach, we will use the equation of motion given:

\[ v^2 = v_0^2 - 2gh \]

where:
- \( v_0 \) is the initial speed of the arrow (70 m/s),
- \( v \) is the speed of the arrow at maximum height (0 m/s, since the arrow stops rising at its peak),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²),
- \( h \) is the maximum height we want to find.

### Step 1: Substitute the known values into the equation

At maximum height, the speed \( v \) is 0 m/s. Thus, we can substitute \( v = 0 \) and \( v_0 = 70 \) m/s into the equation:

\[ 0^2 = (70)^2 - 2(9.8)h \]

### Step 2: Simplify the equation

Calculating \( (70)^2 \):

\[ 0 = 4900 - 2(9.8)h \]

Now, calculate \( 2(9.8) \):

\[ 0 = 4900 - 19.6h \]

### Step 3: Rearrange the equation to solve for \( h \)

Rearranging the equation gives:

\[ 19.6h = 4900 \]

Now, divide both sides by 19.6 to isolate \( h \):

\[ h = \frac{4900}{19.6} \]

### Step 4: Calculate \( h \)

Now we perform the division:

\[ h \approx 250 \]

### Conclusion

The maximum height \( h \) that the arrow will rise to is approximately **250 meters**.