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A stone is thrown vertically upwards with an initial velocity of v what is the distance covered in time 1.5v/g

A stone is thrown vertically upwards with an initial velocity of v what is the distance covered in time 1.5v/g

Grade:9

1 Answers

Arun
25758 Points
4 years ago
Dear Akshat
 
P = v - g * t(1)
t(1)= v/g
 
0² = v² - 2g H
H = v²/2g
So during time t(1), stone will travel = v²/2g = 0.5 v²/g
 
Remaining time(t2) = (1.5 -1) v /g = 0.5 v/g
 
Durung next phase of its downward journey, it starts with velocity =0, and after time( t2 = 0.5 v/g)
Its displacement wil be =
h = 0 *t(2) + (1/2) g* t²
h = (1/2) g* (0.5 v/g)²
h = 0.125 v²/g
 
Hence totak distance covered = (0.5 +0.125) v²/g
= 0.625 v²/g
 
Regards
Arun (askIITians forum expert)

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