from the top of a tower of height 78.4m two stones are projected horizontally with speed 10m/s and 15m/ s in opposite directions the distance of separation between them on reaching the ground is
Answers
On reaching the ground, their separation is 100m .
Step-by-Step Explanation:
Tower height = 78.4 m
S = ut + (1/2)at²
The Vertical initial speed is zero
=> 78.4 = (1/2)(9.8)t²
=> t² = 16
=> t = 4
Horizontal Distance by 10 m/s = 10 * 4 = 40 m
Horizontal Distance by 15 m/s = 15 * 4 = 60 m
Distance of separation = 40 + 60 = 100 m
Therefore, On reaching the ground, their separation is 100m.
Given:
Height of the tower = 78.4m
Speed of stone 1= 10m/s
Speed of stone 2= 15m/s
To find:
The distance of separation between the stones on reaching the ground when projected in opposite directions.
Solution:
Since the height of the tower = 78.4m
And the initial vertical speeds of the stones is 0
By using the formula S = ut + 1/2at^2
78.4 = (1/2)(9.8)t²
t² = 16
t = +4, -4
Since -4 is not possible, therefore t=4 sec
Horizontal distance covered by the stone 1 when projected with a speed of 10m/s = 10*4 = 40m
Horizontal distance covered by the stone 2 when projected with a speed of 15m/s = 15*4 = 60m
Since they were projected in opposite directions, the separation between them is:
40m + 60m = 100m
Therefore, the distance of separation between the stones on reaching the ground is 100m.