Question

A baseball is thrown vertically upward at a rate of 85 feet per second from an initial height of 6 feet. Use the projectile formula to determine when the ball will hit the ground.

Answer 1

Answer:

t1= 5.35

Step-by-step explanation:

Projectile height formula should be

h= -1/2gt^2+v0t+h0

where

h= height, this question asking time when "ball hit the ground" so height is 0

t= time, we look for this variable

v0= initial velocity= 85 feet/s

h0= initial height = 6 feet

g= gravity, assumed 32 feet/s^2

The calculation will be:

h= -1/2gt^2+v0t+h0

0= -$\frac{1}{2}$ 32 t^2+85t+6

0= -16t^2+85t+6

16t^2 - 85t -6 =0

x=  $\frac{-b±\sqrt{ b^2-4ac}}{2a}$

t= $\frac{(-(-85)±\sqrt{ -85^2-4(16)(-6)}}{2(16)}$

t=  $\frac{(-(-85)±\sqrt{7417}}{2(16)}$

t= $\frac{85±\sqrt{7417}}{32)}$

t= $\frac{85±\ 86.12}{32}=$

t1= $\frac{85+\ 86.12}{32}=$

t1= 5.35

t2= $\frac{85-\ 86.12}{32}=$

t2= -0.035

The answer can't be minus, so its t1