NEET Class neet Answered
a body dropped from top of the tower fall through 40m during the last two seconds of its fall the height of the tower
Asked by frozen.queen2488 | 01 Jun, 2022, 18:23: PM
Expert Answer
Let A be the point at Top of tower of height h . Let C be the point at bottom of tower.
Let B be the point where the body reaches 2 second before reaching the bottom of tower .
At the point B , velocity v of body is given as
v = g × (t-2 ) .............................. (1)
From B to C , the body has travelled 40 m distance in 2 seconds. hence we have
40 = [ g × (t-2) ] × 2 + (1/2) g 22 ...........................(2)
( equation of motion " S = u t + (1/2) g t2 " is used to get above equation (2) )
By simplifying eqn.(2) , we get
2 g (t-1)= 40
Hence time t is calculated as
t = [ 40 / ( 2 × 9.8) ] +1 = 3.04 s
Total height of the tower is calculated as
h = (1/2) g t2 = (1/2) × 9.8 × 3.04 × 3.04 = 45.31 m
Answered by Thiyagarajan K | 01 Jun, 2022, 20:46: PM
NEET neet - Physics
Asked by cmashudrana | 26 Jul, 2024, 10:51: AM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by upparmanjunath70 | 19 Jul, 2024, 22:26: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by nk4746870 | 15 Jul, 2024, 21:21: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by dshyamala44 | 08 Jul, 2024, 21:05: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by mnedits01 | 03 Jul, 2024, 15:55: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by musamaria321 | 28 Jun, 2024, 20:31: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by 3s6ng9 | 05 Jun, 2024, 23:59: PM
ANSWERED BY EXPERT
NEET neet - Physics
Asked by srabaneebiswal03 | 27 May, 2024, 10:57: AM
ANSWERED BY EXPERT