From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take 2 g ? 10m /s ) (a) 5 : 7 (b) 7 : 5 (c) 3 : 6 (d) 6 : 3
Added by George K.
Step 1
Step 1: Calculate the distance covered in the 3rd second: Given initial velocity (u) = 10 m/s Acceleration due to gravity (g) = 10 m/s^2 Time (t) = 3 seconds Using the formula: distance = ut + (1/2)gt^2 distance = 10*3 + (1/2)*10*3^2 distance = 30 + 45 distance = Show more…
Show all steps
Your feedback will help us improve your experience
Kamlesh Goyal and 95 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s. The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (a) 5 : 7 (b) 7 : 5 (c) 3 : 6 (d) 6 : 3
Kamlesh G.
A particle is thrown vertically upwards with a speed of $50 \mathrm{~m} \mathrm{~s}^{-1}$. The ratio of the distance travelled in the 5 th second to that in the 6 th second is $\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}\right)$ (a) $2: 1$ (b) $5: 3$ (c) $4: 5$ (d) $1: 1$
A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g ? 10 m/s 2 ) (a) 8 m (b) 10 m (c) 12 m (d) 16 m
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD