﻿ A dynamite blast blows a heavy rock starting up with a launch velocity to 160 m/sec. It reaches a height of s = 160t – 16t2 after t sec. The velocity of the rock when it is 256 m above ground on the way up is : Kaysons Education

# A Dynamite Blast Blows A Heavy Rock Starting Up With A Launch Velocity To 160 M/sec. It Reaches A Height Of s = 160t – 16t2 after t sec. The Velocity Of The Rock When It Is 256 M Above Ground On The Way Up Is

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## Question

### Solution

Correct option is

96 m/s

We now find values of t for which s(t) = 256. So

160t – 16t2 = 256

⇒ 16(t2 – 10t + 16) = 0

⇒ (t – 2) (t – 8) = 0

⇒ t = 2, t = 8

So          v(2) = 160 – 32.2 = 96

v(8) = 160 – 256 = –96.

So the velocity on the way up in 96 m/s.

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