Abaseball 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.

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= - 32 t^2+85t+6

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

16t^2 - 85t -6 =0

x=  

t=

t=  

t=

t=

t1=

t1= 5.35

t2=

t2= -0.035

The answer can't be minus, so its t1

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation: