Question

A ball is thrown straight up from the top of a tower that is 280 ft high with an initial velocity of 48 ft/s. The height of the object can be modeled by the equation s(t) = -16t^2 + 48t + 280.
Determine the time(s) the ball is lower than the building. Express your answer in interval notation. In two or more sentences, describe your solution method.

Answer 1

Answer:

Interval where height of ball less than building = (3, 5.94]

Explanation;

 The equation of the ball thrown from 280 ft height is given by, $-16t^2+48t+280$, where t is the time.

This ball is thrown up so initially the displacement is 280 ft, then it increases and then it reduces after reaching maximum height. When coming down it reaches at 280 ft height after a certain time. After that time the ball is at a height than building.

So we have height = s(t) = 280 ft

  $s(t)=-16t^2+48t+280=280\\ \\ -16t^2+48t=0\\ \\ t^2-3t=0\\ \\ (t-3)t=0$

So, t = 0 or t =3.

We need the 3 seconds value, after that the the ball is lower than the building.

Time to reach ground is given by substituting s(t) =0

  $-16t^2+48t+280=0\\ \\ -2t^2+6t+35=0$

  t = −2.9441 ( not possible)  or t = 5.9441

So at time t = 5.94 it reaches ground.

Interval where height of ball less than building = (3, 5.94]