two balls A and B are thrown simultaneously vertically upwards with a speed of 20 metre per second from the ground and the downwards s from a height of 40m with the same speed and along the same line of motion at what point does the ball collide?
two balls A and B are thrown simultaneously vertically upwards with a speed of 20 metre per second from the ground and the downwards s from a height of 40m with the same speed and along the same line of motion at what point does the ball collide?
ANswers
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.=(40-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.
Like if satisfied
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.=(40-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.
Like if satisfied
