Question

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​

Answer 1

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

Answer 2

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