The secret lies up

Classical Mechanics Level 3

A particle is thrown upward with a velocity of $20\text{ m/s}$ . Find the minimum distance covered by the particle in a time interval of 1 second.

Take gravity as $10\text{ m/s}^2$ .

2.5\text{ m}

5\text{ m}

20\text{ m}

1.25\text{ m}

2 solutions

A Former Brilliant Member
May 30, 2016

minimum distance at highest point therefore answer is 2 1/2 g*(1/2)^2

Can u explain it a bit clearly

Sahasra Ranjan - 5 years ago

Patrick Feltes
Jun 17, 2016

The minimum distance traveled in the one second interval must be symmetric around the highest point. This is because the velocity of the particle is zero at its highest point, allowing for the least amount of distance to be traveled. So, we must look at the period of 0.5 seconds before the highest point and 0.5 seconds after the highest point.

0.5 seconds before

$\Delta y = 5(0.5)-(0.5)(10)(0.5)^2$

$\Delta y = 1.25m$

Since we have symmetry about the center, the vertical distance traveled by the particle in the 0.5 seconds after will also be 1.25 m. Thus, the answer is

$1.25*2=2.5m$

Very very misleading question.

Greg Grapsas - 1 year, 9 months ago

The secret lies up

2 solutions

0 pending reports