Atennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 130 feet per second. what is the maximum height, in feet, the ball will attain

ymax = 266.06 ft : maximum height

Explanation:

when an object is thrown vertically up or down from a certain height H disregarding any type of friction with the air or any other obstacle, it is a uniformly accelerated rectilinear motion in which the acceleration coincides with the value of gravity.

On the surface of the Earth, the acceleration of gravity can be considered constant, directed downwards.

To study the vertical launch movement we will normally use a reference system whose coordinate origin is at the foot of the vertical of the point from which we launch the body and consider the positive direction of the axis and pointing up.

vf²=v₀²+2*g*Δy Formula (1)

Where:  

Δy: vertical displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration  due to gravity in m/s²

Problem development

Data

v₀= 130 ft/s

y₀ = 2 ft

g= 32 ft/s²

In vertical launch of an object up, the acceleration has a sign opposite to the movement .

We apply the formula (1):

vf²=v₀²-2*g*Δy

At maximum height vf = 0

0= (130)²-(2)*(32)*(Δy)

64*(Δy)=  (130)²

Δy= (130)²/64

Δy= (130)²/64

Δy=264.06 ft

Δy = ymax-y₀    ,   ymax = maximum height

ymax= Δy+y₀

ymax= 264.06 ft+2ft

ymax = 266.06 ft

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

spare some change and give it to the poor

explanation:

plz : ((