Given the acceleration, initial velocity, and initial position of a ball thrown vertically upward, how long does it take for it to reach its maximum height? what is the maximum height?

ymax=  y₀ + v₀²/2g

Explanation:

The equations of uniformly accelerated rectilinear motion of upward (vertical) for the y axis are :

vfy = v₀y-gt Formula (1)

vfy² = v₀y²-2gΔy   Formula (2)

Where:  

t: time in any position (s)

Δy= y-y₀

y = vertical position in any time (m)

y₀ : initial vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀y = v₀  : total initial speed of the ball

y₀ : initial vertical position of the ball

g  :acceleration due to gravity

Time (t) calculation for the ball to reach maximum height

We apply the formula (1)

vfy =  v₀y-gt

When the ball reaches its maximum height (h), vy = 0:

0= v₀-gt

v₀ = gt

Calculation of the maximum hight that reaches the ball

When the ball reaches its maximum height (ymax), vy = 0:

We apply the formula (2)

vfy²= v₀y²-2gΔy

0= v₀²-2gΔy

2gΔy  = v₀²

Δy= v₀²/2g    ,Δy= ymax-y₀

ymax-y₀ = v₀²/2g

ymax=  y₀ + v₀²/2g

d.7.5cm is your answer

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.