concept of physics of ninth class from gravitational chapter

फिजिक्स में free fall जुड़ा महत्वपूर्ण concept

जब कोई ऑब्जेक्ट को किसी initial velocity से भेजा जाता है और ;उसमे लगा समय =t

और जब ये वापस surface  पर आता तो   लगने वाला समय =t"

तो क्या ग्रेविटी opposite एंड similiar में सामान समय लगाती है =t+t"/2

ले कर चलेंगे ,

Question 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 (c) its position after 4 s.

Class 9 - Science - Gravitation Page 144

Time to reach Maximum height ,

t = 6/2 = 3 s.

v = 0 (at the maximum height )

a = - 9.8 m s-²

◆ a) Using, v = u + at, we get 

0 = u - 9.8 × 3

or, u = 29.4 ms-¹

Thus, the velocity with which it was thrown up = 29.4ms-¹

◆ b) Using, 2aS = v² - u², we get

S = v²- u²/2a 

= 0 - 29.4 × 29.4/- 2× 9.8

= 44.1 m

Thus, Maximum height it reaches = 44.1 m.

◆ c) Here, t = 4s. In 3 s, the ball reaches the maximum height and in 1 s it falls from the top.

Distance covered in 1 s from maximum height, 

S = ut + 1/2at ² 

= 0 + 1/2 × 9.8 × 1 

= 4.9 m