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Question (Class 11 – Kinematics):
A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (a) the maximum height reached (b) the time taken to return to the thrower. m karim physics numerical book solution class 11
Solution (from M Karim manual):
Step 1: Data Collection
Step 2: Formula for Height Using ( v^2 = u^2 + 2as ): ( 0 = (49)^2 + 2(-9.8)(s) ) ( 19.6s = 2401 ) ( s = \frac240119.6 = 122.5 ) meters.
Step 3: Formula for Time of Flight ( T = \frac2ug = \frac2 \times 499.8 = 10 ) seconds. Let us replicate the quality you should expect
Final Answer: Max Height = 122.5 m; Time of flight = 10 s.
First, a quick clarification: M. Karim is often confused with other authors (like H.C. Verma). While H.C. Verma focuses on conceptual theory, M. Karim’s book is exclusively dedicated to numericals. A ball is thrown vertically upwards with a
A: Absolutely. The numerical style of M Karim is slightly harder than JEE Main, making it perfect practice.