Questions 1

Classical Mechanics Level 3

A small 1.35 𝑘𝑔 mass is launched from the top of a cliff at an angle of 15.9° above the horizontal. When the mass reaches the ground 4.33 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 later, its velocity is directed at 34.4° below the horizontal. What is the speed of the mass when it reaches the ground? Ignore air resistance


The answer is 54.1.

Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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1 solution

Siddharth Tiwari
Aug 16, 2014

Let projecting angle be A and angle at which mass hit the ground be B, and let initial velocity be u and final velocity be v, so,

(u * cos A) = (v * cos B), from this you will get (u / v).

Next use formula, S = u * t - 0.5 * g * t^2 for the y-direction motion.

then substitute the value (u / v) and the given values and you will get v = 54.125 m/s

0 Helpful 0 Interesting 0 Brilliant 0 Confused

take g = 10 m/s^2

Siddharth Tiwari - 6 years, 10 months ago

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Does it matter if g is taken to be 9.8m/s² ? Then the answer will be near 54.03m/s . Brilliant recognizes the actual answer±.5%(it may be different, but near this limit) to be acceptable. But its good for the Asker to mention the values that should be taken for the Solution. Because sometimes it makes a quite huge difference. For example, Einstein's Time Dilation, t=t•/(1-v²/c²) Here, you will find different answer if c is taken to be 3×10^8m/s and if the actual value is taken, which is 2.99×10^8m/s.

Muhammad Arifur Rahman - 6 years, 6 months ago

You should've clarified your solution. Its a Level 3 question, isn't it? You have to explain at least the concept of your equations. Like how you get ucosA=vcosB. It may take up a little of your time. But it will help our Brilliant to be more precise and simplified.

Muhammad Arifur Rahman - 6 years, 6 months ago

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First of all I would like to apologize for late reply. And Arifur Rahman I got the answer using g=10 m/s^2, that is why I mentioned it. And as for the relation that I used ucosA=vcosB, this relation is quite simple to deduce. For this projectile motion only acceleration is in the downward direction (acceleration due to gravity) and since there is no acceleration in horizontal direction (also air friction is not considered) so the velocity in the horizontal direction will be constant i.e. u cosA=v cosB

I hope this solves your doubt.

Siddharth Tiwari - 5 years, 10 months ago

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