Question

from a point a 80m above the ground particle is projected vertically upwards with velocity of 29.4 metre per second. Five seconds later another particles is dropped from a point B 34.3m vertically below A. Determine when and where one overtake the other (g=9.8)

Answer 1

Answer:

Two particles meet at a height of 1.6 m from the ground after 8 second of projection of particle first particle.

Explanation:

Given:

Height at which the  first particle is projected  h=80 m

initial velocity of the particle(1)=29.4 m/s

Let consider the motion of first particle

After 5 second according to equation of motion under gravity $y_1$ be its displacement from initial position which is given by

$y_1=ut-\dfrac{gt^2}{2}\\y_1=29.4\times 5-\dfrac{9.8\times 5^2}{2}\\\\y_1=24.5\ \rm m$

The velocity of the particle

$v=u-gt\\v=29.4-9.8\times 5\\v=-19.6\ \rm m/s$

Now we will use relative motion concept

Now after 5 s when the particle (2) is dropped the relative acceleration between these is 0 as on both the particles gravity is acting in downward direction. The relative velocity between them is 19.6 m/s in downward direction . Let t be the time after which they meet The relative distance between them is (34.3+24.5)=58.8 m

Now

$58.8=19.6\times t\\t=3\ \rm s$

The distance covered by second particle in 3 s

=$\dfrac{gt^2}{2}\\\\=\dfrac{9.8\times 3^2}{2}\\\\=44.1\ \rm m$

So they meet at a height of (80-34.3-44.1)=1.6 m

Answer 2

Answer:

Explanation
Let at time t, they reach level C.
For A:- -(h + 34.3) = 29.4t - (1/2)gt
2
...(i)
For B:- -h = −
2
1
​
g(t−5)
2
...(ii)
(ii) - (i) ⇒ 34.3
= -29.4t + (1/2)g [t
2
−(t−5)
2

⇒ 7 = -6t + (2t - 5) 5 ⇒ t = 8s
Now from (ii), h = (1/2) (9.8) (8−5)
2
- 44.1 m.
From figure,
H = 80 - 34.3 = 45.7 m.
So finally, h
1
​
= H - h = 45.7 - 44.1 = 1.6