    Question

# A stone is dropped from a tower of height 100 m. At the same time, another stone is thrown upwards at a speed of 25 m s−1. When and where the two stones would meet? Take g=10 m s−2

A
Time =2.5 s ; Height =40 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Time =4 s ; Height =80 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Time =6 s ; Height =50 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Time =7.5 s ; Height =100 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B Time =4 s ; Height =80 mGiven: The height of the tower, h=100 m Initial velocity of the second stone, u2=25 m s−1 g=10 m s−2 Initial velocity of the first stone, u1=0 m s−1 Let t be the time instant at which the two stones would meet/hit each other. For the falling stone, the distance covered: s1=u1t+12gt2 ⟹s1=0+12×10t2=5t2 ------(1) For the rising stone, the distance covered: s2=u2t−12gt2 ⟹s2=25t−12×10t2=25t−5t2 But, s1+s2=100 ⟹5t2+25t−5t2=100 ⟹25t=100⟹t=4 s ----- (2) Substitute (2) in (1): s1=5×(4)2=5×16=80 m --- (3) ∴ The two stones would meet each other at t=4 s, and at a height of 80 m.  Suggest Corrections  0      Similar questions  Explore more