()A ball is dropped from the top of a tower of
height 200m & at the same time another ball is
projected from the base of the tower with a
velocity of 40m/s. When the two ball will meet.
(a) 5s
(b)65
(c)2s
(d)3s
In the above question at what distance from the top they will meet.
Answers
Answer: The answer is 5 sec
Explanation: According to the question
time taken = 5sec and distance from the top they will meet is 125 m
let after t time both will meet.
distance travelled by first ball, s = 1/2 gt²
[ as initial velocity of ball, u = 0 ]
distance travelled by 2nd ball, s = 40t - 1/2 gt²
[ initial velocity of second ball is 40m/s ]
height of tower = 200
⇒distance travelled by first ball + distance travelled by 2nd ball = 200 m
⇒1/2 gt² + 40t - 1/2 gt² = 200
⇒t = 200/40 = 5sec
again, distance from the top they will meet, s = 1/2 gt²
= 1/2 × 10 × (5)² [ g = 10m/s² ]
= 125 m
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