Question

an archer shoots an arrow horizontally that hits a target 100m away. A bystander times the flight of the arrow at 2.2s. a) what was the initial velocity of the arrow? b) How far did the arrow drop in the vertical direction?

Answer 1

The archer is fired horizontally.

It hits the target which is at 100 m distance.

The horizontal velocity of the archer is always constant for a projectile in this situation.

The total time of flight is given as 2.2 s

Let the initial velocity is u .

Hence the target distance = ut

                 ⇒ 100 m = u×2.2 s

                   ⇒u = 100÷2.2 m/s

                           =45.45 m/s

We are asked to calculate the vertical distance travelled.

The vertical distance travelled is calculated as-

                                      $s= ut +\frac{1}{2} at^2$  [ s is the distance and a is the acceleration]

                                      $s = 0*2.2 -\frac{1}{2} g[2.2]^2$ [g is te acceleration due to gravity]

                                      $s=\frac{1}{2} 9.8*[2.2]^2$  [here we have taken only magnitude]

                                             =23.716 m       [ans]