Question

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-5t^2 from the ground after a flight of 't' seconds.Find the time taken by the object to touch the ground.

Answer 1

Answer:

4 seconds

Step-by-step explanation:

Since the height s is relative to the ground, the object touches the ground when s = 0.  So...

12 + 17t - 5t² = 0

=> 5t² - 17t - 12 = 0

=> t = ( 17 ± √( 17² - 4 × 5 × -12 ) ) / ( 2 × 5 )

     = ( 17 ± √( 289 + 240 ) ) / 10

     = ( 17 ± √529 ) / 10

     = ( 17 ± 23 ) / 10

     = -6/10  or  40/10

     = -3/5   or   4

Since time moves forwards, we must have t > 0, so t = 4.

Answer 2

Time taken by object to touch the ground=4sec