An object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height.It is at a height of S=12+17t-5 $t^{2}$ from the ground after a flight of 't' seconds. ____ is the time taken by the object to touch the ground.

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Since the height s is relative to the ground, the object touches the ground when s = 0. So...
12 + 17t - 5t² = 0
Or, 5t² - 17t - 12 = 0
Or, t =

\frac{(17 \pm \sqrt (17 ² - 4 \times 5 \times - 12))}{(2 \times 5)}

\frac{(17 \pm \sqrt (289 + 240))}{10}

\frac{(17 \pm \sqrt 529)}{10}

\frac{(17 \pm 23)}{10}

\frac{- 6}{10}

\frac{40}{10}

= -

\frac{3}{5}

or 4
Since time moves forwards, we must have t > 0, so t = 4.
4sec is the time taken by the object to touch the ground.

Watch 3-min video & get full concept clarity