Question

Billy-joe stands on the talahatchee bridge kicking stone into the water below.

a) If Billy-Joe kicks a stone with a horizontal velocity of 3.5 m/s, and it lands in the water a horizontal distance of 5.4 m/s from where Billy-Joe is standing, what is the height of the bridge?

Answer 1

11.6 m

Step-by-step explanation:

First of all, we need to calculate the time of flight of the stone. This can be done by analyzing the horizontal motion only, which is a uniform motion with constant velocity
`v_x = 3.5 m/s`. The time of flight is:

`t=(d)/(v_x)`

where d = 5.4 m is the horizontal distance covered by the stone. Substituting,

`t=(5.4)/(3.5)=1.54 s`

Now we can analyze the vertical motion, which is a uniform accelerated motion with constant acceleration g = 9.8 m/s^2 downward. The vertical distance covered (which is the height of the bridge) is

`h=ut+(1)/(2)gt^2`

where

u = 0 is the initial vertical velocity

t = 1.54 s is the time of flight

Substituting, we find

`h=0+(1)/(2)(9.8)(1.54)^2=11.6 m`