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

$\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f$

$\frac{1}{2}(50)^2 + 9.8(10) = \frac{1}{2}v_f^2 + 0$

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

$F = ma$

$F = 0.050 \times 8986.7$

$F = 449.3 N$