A ball is projected vertically up from the foot of a tower height 100m with avelocity of 40m m/s at the same instant another ball is dropped from the top of the tower when and where do they meet each other
Answers
Let the ball dropped down be b1. So the ball thrown up is b2.
b1 => d = 100-x
g = 9.8 m/s sq.
u = 0
We know, s= ut + 1/2at sq.
Putting values,
100-x = 4.9t sq. ....(1)
b2 => d = x
g = -9.8 m/s sq.
u = 25 m/s
We know, s= ut + 1/2at sq.
Putting values,
x = 25t -4.9t sq. ....(2)
Adding (1) and (2),
100-x +x = 4.9t sq. + 25t - 4.9t sq.
t = 4 secs.
Puuting t = 4 in (1),
100-x = 4.9t sq.
100-x =19.6
x = 80.4 m
So, they meet at 80.4 m from ground after 4 seconds.
Concept:
The second equation of motion,
s = ut + (1/2) at²
where s is the distance traveled, u is the initial velocity, t is the time and a is the acceleration.
Given:
The height of the tower is 100m.
Object projected vertically upward with a velocity of 40 m/s.
Find:
When and where do both the balls meet each other.
Solution:
Let the distance covered by the ball thrown upwards to meet the other be x. So, the distance traveled by the other ball is (100-x) m.
For ball 1, using the second equation of motion,
s = ut + (1/2) at²
Initial velocity is 0, so, u = 0 m/s
100-x = (1/2) gt²
For ball 2, using the second equation of motion,
s = ut + (1/2) at²
Initial velocity is 40 m/s, so, u = 40 m/s
x = 40t - (1/2) gt²
Adding both the equation,
100-x+x = (1/2) gt² + 40t - (1/2) gt²
100 = 40t
t = 2.5 s
Substituting the value of t = 2.5 s to calculate the position of meet,
x = 40t - (1/2) gt²
x = 40(2.5) - (1/2) g(2.5)²
x = 100 - 31.25 = 68.75 m
Hence, both the ball will meet at 2.5 seconds at the distance of 68.75 m from the ground.
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