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wo balls A and B are thrown simultaneously, A vertically upwards with a speed of 20m/s from the ground, and B vertically downwards from a height of 100 m with the same seed and along the same line of motion. At what points do the two balls collide?

wo balls A and B are thrown simultaneously, A vertically upwards with a speed of 20m/s from the ground, and B vertically downwards from a height of 100 m with the same seed and along the same line of motion. At what points do the two balls collide?

Grade:11

2 Answers

Arun
25750 Points
4 years ago
Case 1:
For the ball A :
initialSpeed=u=20m/s[upwards]=-20m/s
a=10m/s²[downwards]=+10m/s²
time=t=tsec
Distance=Sa meters
From seond equation of motion,
Sa=ut+1/2at²
Sa=-20t+1/2x10xt²
Sa=-20t+5t² --------------equation(1)
Case II:
For ball B:
Initial speed=u=20m/s[downwards]=+20m/s
a=10m/s²[downwards]=+10m/s²
time =t sec
Distance =Sb meters.=(100 – Sa)
From seond equation of motion,
Sb=ut+1/2at²(4-Sa)
=20t+5t² ====[equation2]
Therefore , from equation 1 and 2 we get:
40=-20t+5t²+20t+5t²40
=10t²4=t²
t=2sec
Let us find out distance Sa in 2 sec
Sa=ut+1/2at²
sa=-40+20=-20m
From the ground that is at 20m the two balls will collide.
 
 
Khimraj
3007 Points
4 years ago
U1=20 m/s (upward)
U2=20(downward)
Let X be the distance from the ground where the collision takes place then
Distance from top of tower up to the collision point =40-X
If we suppose that collision took place after time T Then
X=20T-1/2×g×T^2……(1)(upward motion)
40-X=20T+1/2g×T^2….(2)(downward motion)
Adding (1) and (2) we get
40=40T
Which gives T=1 second
Putting value of T in (1) we get
X=20–1/2 ×9.8×1 ( taking :g= 9.8 m /s^2)
X=20–4.9=15.1 metre
So collision takes at 15.1 metres from ground or 40–15.1=24.9 metres from top.

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