Question

A 50g arrow is fired vertically downward to a target 10m below. If it was released from the bow with a speed of 50m/s, and it lodges itself 15cm into the ground determine the average force that the arrow exerts on the ground ignore air resistance

Answer 1

Speed of the arrow just before it will hit the ground can be calculated by energy conservation

`(1)/(2)mv_i^2 + mgh_i = (1)/(2)mv_f^2 + mgh_f`

`(1)/(2)(50)^2 + 9.8(10) = (1)/(2)v_f^2 + 0`

by solving above equation

`v_f = 51.9 m/s`

now this will lodge itself 15 cm into the ground

so after lodging itself to 15 cm the speed of arrow becomes zero

now we will have

`v_f^2 - v_i^2 = 2a d`

`0 - 51.9^2 = 2(a)(0.15)`

`a = -8986.7 m/s^2`

Now by Newton's II law net force is given as

`F = ma`

`F = 0.050 * 8986.7`

`F = 449.3 N`