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 +(1)/(2) at^2` [ s is the distance and a is the acceleration]

`s = 0*2.2 -(1)/(2) g[2.2]^2` [g is te acceleration due to gravity]

`s=(1)/(2) 9.8*[2.2]^2` [here we have taken only magnitude]

=23.716 m [ans]