Question

An arrow of mass 0.10 kg was fired horizontally from a height of 1.5 m by an archer who exerted a force of 350 N on the bowstring and pulled the string back 0.70 m: How far from the archer did the arrow land?

Answer 1

The arrow lands 12.2 m away from the archer.

Step-by-step explanation:

We can start by using conservation of energy, where the initial potential energy of the arrow is converted into kinetic energy as it is fired. We can assume that there is negligible air resistance.

Initial potential energy = mgh = 0.10 kg x 9.81 m/s^2 x 1.5 m = 1.47 J

The work done by the archer in pulling the bowstring is equal to the change in potential energy:

Work = force x distance = 350 N x 0.70 m = 245

Therefore, the kinetic energy of the arrow when it is fired is:

Kinetic energy = Work = 245 J

Using the formula for kinetic energy:

Kinetic energy = 0.5mv^2

245 J = 0.5 x 0.10 kg x v^2

v^2 = 490 m^2/s^2

v = √(490) = 22.1 m/s

The horizontal distance traveled by the arrow is equal to the horizontal component of its velocity multiplied by the time it takes to hit the ground. The time can be found using the vertical displacement of the arrow and the acceleration due to gravity:

h = 1/2gt^2

1.5 m = 1/2 x 9.81 m/s^2 x t^2

t = √(0.306) = 0.554 s

The horizontal distance is:

d = vt = 22.1 m/s x 0.554 s = 12.2 m

Therefore, the arrow lands 12.2 m away from the archer.