Question

An object is thrown upward from the top of a building with an initial velocity of 48 feet per second. The height h of the object after t seconds is given by the quadratic equation When will the object hit the ground?

Answer 1

Answer:

5 secs

Step-by-step explanation:

The height of the object is given as:

$h = -16t^2+48t+160$

The moment when the object hits the ground, the height of the object will be 0 m.

Hence:

$0 = -16t^2+48t+160$

Solving this for t:

Divide through by 16 and move all parameters to the left hand side:

$t^2-3t-10 = 0$

$t^2 - 5t + 2t - 10 = 0\\\\\\t(t - 5) + 2(t - 5) = 0\\\\\\(t + 2)(t - 5) = 0\\\\\\=> t = -2, 5$

Since time cannot be negative, time, t = 5 secs.

The object will hit the ground after 5 secs.