Question

Two balls are thrown simultaneously, one vertically upward from the ground and the other downward from a height of 40 m. If both are thrown with same velocity, at what height from the ground will they collide?(g=9.8 m/s2).

Answer 1

Dear student,
As you have not specified initial velocity. We can't solve the problem without any clues to proceed.
Given :Let me assume initial velocity=u=20m/s[from Inter xi physics book]
a=9.8m/s²
For ball A:
let x be the height.
From equations of motions,
s=ut-1/2gt²
x=20t-(1/2)x9.8t²-----------(1)
for ball B;
a=-9.8m/s²
u=-20m/s
h=-(40-x)
a=-9.8m/s²
using s=ut+1/2at²
we get,
-(40-x)=-20t-1/2x9.8xt²
40-x=20t+4.9t²------------(2)
By adding equation 1 and 2, we get
x=20t--4.9t²
40-x=20t+4.9t²
+____________

40=40t
_____________

40=40t
t=1sec
substituting value of t in equation 1, we get
x=20x(2) -4.91*1
x=15.1m
∴ At a height of 15.1m above the ground, two balls will collide.