Question

An object is thrown upwards with an initial velocity of 16 ft/sec from a building with a height of 96 feet. It is at a height of S = 96 + 16t² – 16t² from the ground after a flight of 't' seconds. Find the time taken by the object to touch the ground.​

Answer 1

Answer:

When the Object touches the Ground

It's Distance = 0 (Find Velocity = 0)

∴ 12 + 17t - 5t² = 0

= 5t² - 17t + 12 = 0

= t (5t - 12) -1 (5t - 12) = 0

= t (5t - 12) -1 (5t - 12) = 0

= (5t - 12) (t - 1) = 0

∴ (5t - 12) = 0          (t - 1) = 0

= 5t = 12                  t = 1

= t = 12/5

Answer 2

Given:

Initial velocity of the object $=16ft/sec$

Height of building $=96ft$

Distance at any point of the trajectory $=96+16t-16t^{2}$

To find:

Time taken by the object to reach the ground.

Solution:

We have been given that the object thrown from the top of the building of height $96ft$ with an initial velocity of $16ft/sec$ reaches the ground.

We have also been given the equation $96+16t-16t^{2}$ which is the height of the object from the ground at any time t.

Hence,

On reaching the ground, the distance of the object from the ground is obviously 0. Therefore on the ground $S=0$.

$96+16t-16t^{2}=0$

$-16(t^{2}-t-6 )=0$

$t^{2}-t-6=0$

$t^{2}-3t+2t-6=0$

$t(t-3)+2(t-3)=0$

$(t+2)(t-3)=0$

$t=-2,3$

Time can never be negative, Hence $t=3s$

Final answer:

Hence, the object reaches the ground in $3sec.$