A ball is thrown vertically upwards from a height of 8 ft with an initial velocity of 30 ft per second. How high will the ball go? Note that the acceleration of the ball is given by a(t)= -32 feet per second per second a) 50.1875 ft
b) 18.5469 ft
c) 53.1875 ft
d) 22.0625 ft
e) 34.1875 ft

c)

Step-by-step explanation:

u = Initial velocity = 30 ft/s

s = Displacement

v = Final velocity = 0

a = Acceleration due to gravity =

From kinematic equations we have

Distance the ball will reach from the height from which the ball is thrown is .

Distance of the ball from the ground is .

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.