Q.45 Bhishma Pitamah was on whose side in the battle of Mahabharata?(a) Kaurava(b) Pandava(c)Not in any party
Answers
Answer:
karavas
Explanation:
according to his father promise he had to protect hastinapur
Answer:
nice to meet u rahul
which class are u
Explanation:
Answer:-
\small \bf \underline{Given:}
Given:
★★ Height at which the ball is allowed to fall from the top of the tower = 200m
★★ Velocity at which the other ball is thrown vertically upwards from the bottom of the tower = 40 m/s
\small \bf \underline{To \: find:}
Tofind:
We need to find when and where does the two balls meet
\small \bf \underline{Solution:}
Solution:
Let the ball meet at height h from the ground
Therefore, the dropped ball travels a height = 200-h
Distance travelled by the ball thrown vertically upward = h
Let t be the time at which they meet
The height travelled by the dropped ball
\bf200 - h = 0.5 \times g \times t \: \: \: \: \: \: \: \: \: ........(1)200−h=0.5×g×t........(1)
The height travelled by the ball vertically thrown upwards
\bf \: h = 40t - 0.5 \times g \times {t}^{2} \: \: \: \: \: \: \: \: ........(2)h=40t−0.5×g×t
2
........(2)
Adding equation (1) and (2) we get
200 = 40t
\bf \implies \: t = \frac{200}{40}⟹t=
40
200
\bf \implies \: t = 5s⟹t=5s
Substituting the value of t in equation (1), the equation stands as follows:
\bf \: 200 - h = 4.9 \times {5}^{2}200−h=4.9×5
2
\bf \implies \: h = 200 - 4.9 \times 25⟹h=200−4.9×25
\bf \implies \: h = 200 - 122.5⟹h=200−122.5
\bf \implies \: h = 77.5m⟹h=77.5m
Answer: The two balls meet at a height of 77.5m from the ground after