Questions: A ball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=64+48t-16t^2. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down?

A ball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=64+48t-16t^2.
(a) After how many seconds does the ball strike the ground?
(b) After how many seconds will the ball pass the top of the building on its way down?

Transcript text: A ball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. The distance $s$ (in feet) of the ball from the ground after $t$ seconds is $s=64+48 t-16 t^{2}$. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down?

Solution

Solution Steps

Step 1: Understand the Problem

We are given the equation for the distance $ s $ of a ball from the ground after $ t $ seconds: \[ s = 64 + 48t - 16t^2 \] We need to find: (a) The time when the ball strikes the ground. (b) The time when the ball passes the top of the building on its way down.

Step 2: Solve for When the Ball Strikes the Ground

The ball strikes the ground when $ s = 0 $. Set the equation to zero and solve for $ t $: \[ 0 = 64 + 48t - 16t^2 \] Rearrange the equation: \[ 16t^2 - 48t - 64 = 0 \] Divide the entire equation by 16 to simplify: \[ t^2 - 3t - 4 = 0 \] Factor the quadratic equation: \[ (t - 4)(t + 1) = 0 \] The solutions are $ t = 4 $ and $ t = -1 $. Since time cannot be negative, the ball strikes the ground at $ t = 4 $ seconds.

Step 3: Solve for When the Ball Passes the Top of the Building on Its Way Down

The ball passes the top of the building when $ s = 64 $ on its way down. We need to find the second time $ t $ when $ s = 64 $: \[ 64 = 64 + 48t - 16t^2 \] Simplify the equation: \[ 0 = 48t - 16t^2 \] Factor out $ 16t $: \[ 16t(3 - t) = 0 \] The solutions are $ t = 0 $ and $ t = 3 $. The ball passes the top of the building on its way down at $ t = 3 $ seconds.