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
Given : 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
To Find : the distance of separation between them on reaching the ground
Solution:
Height = 78.4 m
Thrown horizontally
hence vertical speed = 0
S = ut + (1/2)at²
S = height = 78.4 m
t = ?
u = 0
a = g = 9.8 m/s²
78.4 = 0 + (1/2)(9.8)t²
=> t² = 16
=> t = 4
Horizontal Speed = 10 m/s and 15 m/s
Horizontal Distance = 10* 4 = 40 m & 15 * 4 = 60 m
As thrown in opposite direction.
distance of separation between them on reaching the ground is = 40 + 60 = 100 m
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