Answer:
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
