Question

An object is thrown upwards with an initial velocity of 17 metre per second from a building with 12 metre height it is at a height of S=12+17t-5t^2 from the ground after a flight of t seconds find the time taken by the object to touch the ground

Answer 1

$v = \frac{ds}{dt} (using \: calculus) \\ therefore \: by \: solving \: we \: get \: \\ v = \frac{d(12 + 7t \times - 5 {t}^{2}) }{dt} = 17 - 5t \\ by \: soving \\ \: v = u + at \\ we \: get \: \\ t = 2sec$

Answer 2

Answer:

After 4 seconds the object will touch the ground.

Step-by-step explanation:

The height of object given by a function

$S=12+17t-5t^2$

where, t is time in seconds.

The height of object is 0, if it touch the ground.

$0=12+17t-5t^2$

$0=12+20t-3t-5t^2$

$0=4(3+5t)-t(3+5t)$

$0=(3+5t)(4-t)$

$t=4,-\frac{5}{3}$

The time can not be negative. Therefore the value of  t is 4 and after 4 seconds the object will touch the ground.