A particle is projected vertically upwards from a point O with initial speed 12.5 m s−1. At the same instant another particle is released from rest at a point 10 m vertically above O. Find the height above O at which the particles meet.?

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A particle is projected vertically upwards from a point O with initial speed 12.5 m s−1. At the same instant another particle is released from rest at a point 10 m vertically above O. Find the height above O at which the particles meet.?

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Answer 1 · 2018-04-09T08:22:35

The height is $= 6.86 m$ $=6.86m$

Explanation:

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The height where the particles will meet is $= l m$ $=lm$

The distance travelled by the particle from the bottom is $= l m$ $=lm$

The distance travelled by particle from the top is $= (10 - l) m$ $=(10-l)m$

Apply the equation of motion,

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

The time when they will meet is $= t s$ $=ts$

For the particle from the bottom

$l = 12.5 \times t - \frac{1}{2} \times 9.8 \times t^{2}$ $l=12.5xxt-1/2xx9.8xxt^2$

For the particle from the top

$10 - l = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}$ $10-l=0xxt+1/2xx9.8xxt^2$

Therefore,

$10 - (12.5 \times t - \frac{1}{2} \times 9.8 \times t^{2}) = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}$ $10-(12.5xxt-1/2xx9.8xxt^2)=0xxt+1/2xx9.8xxt^2$

$10 - 12.5 \times t + \frac{1}{2} \times 9.8 \times t^{2} = 0 \times t + \frac{1}{2} \times 9.8 \times t^{2}$ $10-12.5xxt+1/2xx9.8xxt^2=0xxt+1/2xx9.8xxt^2$

$10 - 12.5 t = 0$ $10-12.5t=0$

$t = \frac{10}{12.5} = 0.8 s$ $t=10/12.5=0.8s$

So,

$l = 12.5 \cdot 0.8 - 4.9 \cdot {0.8}^{2} = 6.86 m$ $l=12.5*0.8-4.9*0.8^2=6.86m$

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A particle is projected vertically upwards from a point O with initial speed 12.5 m s−1. At the same instant another particle is released from rest at a point 10 m vertically above O. Find the height above O at which the particles meet.?

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