a girl throws a dart with a speed of 12.4 meters per second. the dart hits the board .32 meters below the height from which it was thrown. how far from the board was the girl?

that doesnt really answer my question

t=sqrt(2h/g)

L=v• t

To answer the question, we need to calculate the horizontal distance from the girl to the board. We can use the horizontal motion equation, which states:

distance = speed × time

In this case, the initial speed is given as 12.4 meters per second, and there is no horizontal acceleration. Therefore, the speed remains constant throughout the projectile's motion. We can assume the time taken is the same for both the upward and downward motions.

First, let's find the time it takes for the dart to reach its maximum height. We can use the vertical motion equation:

final velocity = initial velocity + (acceleration × time)

Since the dart reaches its maximum height, its final velocity becomes zero. The acceleration is the acceleration due to gravity, which is approximately 9.8 meters per second squared (m/s^2). Rearranging the equation, we get:

time = (final velocity - initial velocity) / acceleration

Substituting the values, we have:

time = (0 - 12.4) / -9.8
time = 1.2653 seconds (rounded to four decimal places)

Now that we have the time it takes for the dart to reach its maximum height, we can double it to calculate the total time for the dart's flight. This is because the time taken to descend is equal to the time taken to ascend.

total time = 2 × 1.2653
total time = 2.5306 seconds (rounded to four decimal places)

Using the total time, we can now calculate the horizontal distance covered by the dart:

distance = speed × total time
distance = 12.4 × 2.5306
distance = 31.3134 meters (rounded to four decimal places)

Therefore, the girl was approximately 31.3134 meters away from the board.