Question

Calculate its
15. A stone is thrown vertically upward with an initial velocity of
40 m/s. Taking g = 10 m/s find the maximum height reached
by the stone. What is the net displacement and the total
distance covered by the stone?
16. Calculate the force of gravitation between the earth and the
Sun, given that the mass of the earth = 6 x 1024 kg and of the
Sun = 2 10 kg. The average distance between the two is
1.5 x 101 m.
17. A stone is allowed to fall from the top of a tower 100 m high
and at the same time another stone is projected vertically
upwards from the ground with a velocity of 25 m/s. Calculate
when and where the two stones will meet.
18. A ball thrown up vertically returns to the thrower after 6 s
Find
(a) the velocity with which it was thrown up.
(b) the maximum height it reaches, and​

Answer 1

Explanation:

15)

Initial Velocity u=40

Initial Velocity u=40Fianl velocity v=0

Initial Velocity u=40Fianl velocity v=0Height, s=?

Initial Velocity u=40Fianl velocity v=0Height, s=?By third equation of motion

v^2 -u^2=2gs

=2gs0−40^2 =−2×10×s

=−2×10×ss= 160/20

⇒s=80m/s

⇒s=80m/sToatl distance travelled by stone = upward distance + downwars distance =2×s=160m

⇒s=80m/sToatl distance travelled by stone = upward distance + downwars distance =2×s=160mTotal Diaplacement =0, Since the initial and final point is same.

18 (a)

The ball returns to the ground after 6 seconds.

Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s

Let the velocity with which it is thrown up be u

(a). For upward motion,

v=u+at

∴ 0=u+(−10)×3

⟹u=30 m/s

18(b) in the pic ↑

16 and 17 are also in pic ↑

Answer 2

Explanation:

15. Answer: The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

Explanation:

Given that,

Final velocity v = 0

Initial velocity u = 40m/s

We know that,

Using equation of motion

v^2=u^2+2ghv

2

=u

2

+2gh

0-40^2= 2\times 10\times h0−40

2

=2×10×h

The maximum height is

h = 80\ mh=80 m

The stone will reach at the top and will come down

Therefore, the total distance will be

$s = h_{1}+h_{2}s=h$

1

+h 2

s = 80\ m+80\ m = 160 ms=80 m+80 m=160m

The net displacement is

$D = h_{1}-h_{2}D=h$

1 −h 2

D = 80\ m-80\ m= 0D=80 m−80 m=0

Hence, The maximum height is 80 m and the total distance covered by the stone is 160 m and the displacement is zero.

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