CBSE Class 11-science Answered
A ball is allowed to fall from the top of a tower 200 m high , At the same instant another ball is thrown vertically upward from the bottom of a tower with a velocity of 40 m per sec. When and where the two ball meets.??
Asked by ravikantmanchanda | 13 Mar, 2016, 10:29: PM
Expert Answer
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:
For the ball dropped from the top of the tower:
Distance covered by the ball is (200 - h) m
Here u = 0 ; s = (200 - h) m and g = 9.8 ms-2
→ s = ut + 1/2 at2
or, 200 - h = 0 × t + 1/2 at2
or, 200 - h = 4.9t2 ............(Equation 1)
Here u = 0 ; s = (200 - h) m and g = 9.8 ms-2
→ s = ut + 1/2 at2
or, 200 - h = 0 × t + 1/2 at2
or, 200 - h = 4.9t2 ............(Equation 1)
For the ball thrown vertically upwards:
u = 40ms-1 ; s = h and g = -9.8ms-2 (-ve value of g since thrown upwards)
s = ut + 1/2 at2
h = 40 × t + 1/2(-9.8)t2
or, h = 40t - 4.9t2 ............(Equation 2)
Adding equations (1) and (2), we get
200 - h + h = 4.9t2 + 40t - 4.9t2
On solving we get
t = 5 s
Substituting t = 5 seconds in equation 1, we get
200 - h = 4.9 × (5)2
or, h = 200 - 4.9 × 25
= 200 - 122.5
= 77.5 m
Substituting t = 5 seconds in equation 1, we get
200 - h = 4.9 × (5)2
or, h = 200 - 4.9 × 25
= 200 - 122.5
= 77.5 m
Thus, the two balls meet at a height 77.5 m from the ground after 5 s.
Answered by Yashvanti Jain | 15 Mar, 2016, 10:33: AM
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