A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t = 2 s. The gap is found to remain constant. The velocities with which the balls were thrown are (Take g = 10 m s -2 )
Text Solution
Verified by ExpertsA
For first stone. Taking the vertical upwards motions of the first stone up to highest point
Here , 
(At highest point velocity is zero )
Or 
…… (1)
For second stone. Taking the vertical upwards motions of the second stone up to highest point
Here, 
As 
……… (ii)
As per questions
Subtract (i) From (ii), we get
On substituting the given information, we get
Or 
Or 
and 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems