A stone is dropped from the top of a tower
and it strikes with 3 km/hr against the
ground. Another stone is thrown vertically
downwards from the same top of the tower
with a velocity 4 km/hr. Its velocity when
it strikes the ground will be
(1) 7.0 km/hr (2) 5.0 km/hr
(3) 3.5 km/hr (4) 4.0 km/hr
Answers
Answered by
0
Answer:
7.0 km/h
Explanation:
v = 3km/hr = 3*(5/18) = 5/6 m/s
V = u + at
5/6= 0 + 10(t)
t = 1/12
When stone is thrown with initial speed of 4km/h :
U = 4 km/h = 10/9 m/s
V= U + at
V = 10/9 + 10*(1/12)
= 10*[(1/9)+(1/12)]
= 10*7/36
So V = (70/36)* 18/5 = 7 km/h
Answered by
0
Answer:
(2)5.0km/hr
Explanation:
The velocity of the first stone on reaching the ground in m/s is = 3×5/18 = 5/6m/s
so v^2=2gh = (5/6)^2
where h in 2gh is the height of the tower now for the 2nd stone the upward initial speed =
4×5/18 = 10/9 m/s
v^2=u^2+2gh = v^2 = (5/6)^2+(10/9)^2
= 1.929 m/s^2 × (18/5)^2
= 25
V = 5 km/h
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