Mechanics - 13

Classical Mechanics Level 3

The gap shown is just for sake of understanding A stone which is at $A$ is projected vertically upwards. It passes through a fixed point $P$ , four seconds after its projection. After three more seconds it again passes through $P$ but this time it is falling towards ground. Find the velocity of the stone when it passes through $P$ (in $\text{m/s}$ ) ?

Assume $g = 9.8 \text{ m/s}^2$

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The answer is 14.7.

2 solutions

Eric Nordstrom
Apr 4, 2019

$\text{acceleration} = \frac{\Delta v}{\Delta t}$

Use point P as the initial and final cases.

$-9.8 = \frac{(-v_P)-v_P}{3}\\ \frac{2}{3}v_P=9.8\\ v_P=\boxed{14.7 \frac{\text{m}}{\text{s}}}$

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/

Jake Tricole - 2 years ago

It is done!

Eric Nordstrom - 2 years ago

Biswajit Mohanty
Apr 1, 2019

Let ,the stone passes through p at a velocity =v. It reaches again point p with velocity =-v. During this motion the only force acting on the stone is gravitational attraction. So,it has a constant acceleration = g m/s^2. For uniform acceleration the motion , Speed can be expressed as V(final)=V(initial)+gt. g=-9.8m/s^2. V(final)=-V V(initial)=V. t=3sec Substituting values in the equation,we get 2V=9.8×3 -->V=14.7 m/s^2.

Mechanics - 13

The answer is 14.7.

2 solutions

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