Question

A ball is thrown vertically upward with an initial velocity of 32 feet per second. The distance s​ (in feet) of the ball from the ground after t seconds is s equals 32 t minus 16 t squared. ​(a) At what time t will the ball strike the​ ground? ​(b) For what time t is the ball more than 12 feet above the​ ground?

Answer 1

Answer:

(a) Therefore the ball will strike the ground after $\frac12$ s.

(b)Therefore the ball is more than 12 feet above the ground for 1 s.

Step-by-step explanation:

Given that,

a ball is thrown vertically upward with an initial velocity of 32 feet per second.

The distance of the ball from the ground is

$S=32 t - 16t^2$

where s is distance in feet and t is time in second.

(a)

When the ball strike the ground, the height of the ball will be zero.

That is S=0

$\Rightarrow 32t-16t^2=0$

$\Rightarrow 16t(2-t)=0$

$\Rightarrow t=0,2$

Therefore the ball will strike the ground after 2 s.

(b)

At some time t₁ between the ball being thrown upward direction and the maximum height, the ball is at a height of 12 ft.

After attain the maximum height, the ball come down towards ground. So, the ball again attain a height 12 ft after t₂ second.

To find the value of t₁ and t₂, we have to solve the following equation

$32t-16 t^2=12$

$\Rightarrow 32t-16 t^2-12=0$

$\Rightarrow -4(4t^2-8t+3)=0$

$\Rightarrow 4t^2-8t+3=0$

$\Rightarrow 4t^2-6t-2t+3=0$

$\Rightarrow 2t(2t-3)-1(2t-3)=0$

$\Rightarrow (2t-3)(2t-1)=0$

$\Rightarrow t=\frac 12, \frac32$

Therefore the ball is more than 12 feet above the ground for $=(\frac32-\frac12)s$

                                                                                                      =1 s