Question

An archer fires an arrow straight up with a speed of Vo= 46.6 m/s. I found the maximum height reached by the arrow to be 110.8m. For part b, write an expression for the time the arrow is in the air before it returns to launch height.

Answer 1

The expression for the time an arrow is in the air before returning to the launch height is given by t = 2 * (V0 / g), where V0 is the initial velocity and g is the acceleration due to gravity.

Step-by-step explanation:

The expression for the time the arrow is in the air before it returns to launch height can be found by using the kinematic equation for vertically launched projectiles under constant acceleration due to gravity. When the arrow is shot straight up with an initial speed V0, the total time t it spends in the air before returning to the same height is calculated as t = 2 * (V0 / g), where g is the acceleration due to gravity (~9.81 m/s2).

For an initial speed V0 of 46.6 m/s, the total time t the arrow is in the air can be calculated as:

t = 2 * (46.6 m/s / 9.81 m/s2)

This results in the arrow being in the air for approximately 9.5 seconds before it lands back at launch height.