A ball is dropped from a height of 20 meters and at the same time another ball is thrown up from the ground with the speed of 20 m/s. When and where will the balls meet?
A ball is dropped from a height of 20 meters and at the same time another ball is thrown up from the ground with the speed of 20 m/s. When and where will the balls meet?
Let us assume that the two balls meet at a height 'h' after time 't' above the ground.
For the ball dropped from the top of the tower:
Distance covered by the ball is (20 - h) m
Here u = 0 ; s = (20 - h) m and g = 9.8 ms-2
Using the second equation of motion,
→ s = ut + 1/2 at2
or, 20 - h = 0 × t + 1/2 at2
or, 20 - h = 4.9t2 ............(Equation 1)
For the ball thrown vertically upwards:
u = 20 ms-1 ; s = h and g = -9.8ms-2 (-ve value of g since thrown upwards)
s = ut + 1/2 at2
h = 20 × t + 1/2(-9.8)t2
or, h = 20t - 4.9t2 ............(Equation 2)
Adding equations (1) and (2), we get
20 - h + h = 4.9t2 + 20t - 4.9t2
On solving we get
t = 1 s
Substituting t = 1 second in equation 1, we get
20 - h = 4.9 × (1)2
or, h = 20 - 4.9
= 15.1 m
Thus, the two balls meet at a height 15.1 m from the ground after 1 s.
Let us assume that the two balls meet at a height 'h' after time 't' above the ground.
For the ball dropped from the top of the tower:
Distance covered by the ball is (20 - h) m
Here u = 0 ; s = (20 - h) m and g = 9.8 ms-2
Using the second equation of motion,
→ s = ut + 1/2 at2
or, 20 - h = 0 × t + 1/2 at2
or, 20 - h = 4.9t2 ............(Equation 1)
For the ball thrown vertically upwards:
u = 20 ms-1 ; s = h and g = -9.8ms-2 (-ve value of g since thrown upwards)
s = ut + 1/2 at2
h = 20 × t + 1/2(-9.8)t2
or, h = 20t - 4.9t2 ............(Equation 2)
Adding equations (1) and (2), we get
20 - h + h = 4.9t2 + 20t - 4.9t2
On solving we get
t = 1 s
Substituting t = 1 second in equation 1, we get
20 - h = 4.9 × (1)2
or, h = 20 - 4.9
= 15.1 m
Thus, the two balls meet at a height 15.1 m from the ground after 1 s.
