Sections:

Two Dimensional Motion and Vectors, Page 10

Solved Example of Horizontal Projection

  1. A stone is projected horizontally with a velocity of 6 m/s from the bridge at a height 10 m from the river. How far away from the bridge is the stone when it reaches the river? 

    Solution: 

    Given: vx = 6 m/s and height of the bridge = 10 m.

    Horizontal distance, x = v sub x t ----------------- (1),

    where ‘t’ is the time taken by the stone to reach the river by the time it crossed the height of the bridge 10 m. Using the following equation, we can find the time ‘t’.
     
    s = vt + 1/2 at^2


    By substituting the values of s = 10 m into the above equation, we can find time ‘t’.

    -10 = 0 x t + 1/2 (-9.8)t^2; t^2 = 10/4.9; t^2 = 2.04; t = square root of 2.04; t = 1.4s


    Using the value of time from work above, we can calculate the horizontal distance as shown below. 
    x = v sub x t; x = 6 x 1.4; x = 8.4 m


    The stone reaches the river 8.4 m away from the bridge.