Question

An archer shoots an arrow horizontally at a target 11 m away. the arrow is aimed directly at the center of the target, but it hits 56 cm lower. What was the initial speed of the arrow?

Answer 1

Step-by-step explanation:

Given :

Distance in the vertical direction, S = 56 cm

= 0.56 m

Initial velocity in the vertical direction, u = 0 m/s

Acceleration due to gravity, g = 9.81
`m/s^2`

Therefore, in order to find the time required,

`$S = ut + (1)/(2)at^2$`

`$0.56 = o(t) + (1)/(2)(9.81)t^2$`

`$t^2=(0.56 * 2)/(9.81)$`

`$t^2=0.114$`

t = 0.33 seconds

Therefore, the initial velocity of the arrow is given by :

`$v_0=(S)/(t)$` , where S = 11 m and t = 0.33 seconds

`$v_0=(11)/(0.33)$`

`$v_0=33.33\ m/s$`

Thus the initial velocity of the arrow is 33.33 m/s.