An archer fires an arrow at a velocity of #40# #ms^-1# and an angle of #45^o# to the horizontal, from a point #1.5# #m# above the ground. Will the arrow reach a tree #400# #m# from the archer?

1 Answer
Nov 29, 2017

The arrow will travel a little more than #164# #m# before it hits the ground. It will definitely not make it to a tree #400# #m# away.

Explanation:

First find the two components of the initial velocity:

#u_x=ucostheta=40cos45=28.3# #ms^-1#
#u_y=usintheta=40sin45=28.3# #ms^-1#

The horizontal velocity will remain constant, the vertical velocity will change due to the acceleration due to gravity.

In the vertical direction:

#v_y=u_y+a_yt#

The vertical component of the velocity will decrease to #0# #ms^-1# at its maximum height. We can find the time taken to reach that point:

#y=(v_y-u_y)/a_y=(0-28.3)/(-9.8)=2.9 s.

The acceleration due to gravity takes a negative sign because it is in the opposite direction to the initial velocity.

The total time of flight will be twice this, so #2xx2.9=5.8# #s#.

The horizontal distance traveled in this time will be given by

#s=u_xt=28.3xx5.8=164.1# #m#

This is the distance traveled by the time the arrow returns to the level at which it was released, #1.5# #m# above the ground. It will travel some additional distance, and we can calculate that, but it will definitely not reach a tree #400# #m# away.