An arrow is shot vertically upwards with an initial velocity of 14 m/s . Calculate the height reached

potential energy at top = kinetic energy at bottom

m g h = (1/2) m v^2

h = v^2/(2g)

max height is reached when it comes to a stop

v = 14-gt
solve for t when v=0 and use that to find h in

h = 14t - g/2 t^2

Or, use (or derive) a simpler more direct formula without finding t first.

To calculate the height reached by the arrow, we can use the equations of motion. The key equation we need is the equation for vertical displacement (height) when an object is launched vertically upwards:

h = v^2 / (2g)

Where:
h = height reached
v = initial velocity of the arrow
g = acceleration due to gravity (approximately 9.8 m/s^2)

Given:
v = 14 m/s
g = 9.8 m/s^2

Now, we can substitute the values into the equation:

h = (14 m/s)^2 / (2 * 9.8 m/s^2)

h = 196 m^2/s^2 / 19.6 m/s^2

h ≈ 10 m

Therefore, the arrow reached a height of approximately 10 meters.