A ball is thrown upward with initial velocity of the earth's surface 96 feet per second. What is the maximum height reached? A. A ball is thrown upward with initial velocity of the earth's surface 96 feet per second. What is the maximum height reached?

Question

anonymous · Answer

use newton's laws... v=0,u=96,a=-10 use third law you'll get h=v*v/2a

anonymous · Answer

can we use integral for solved this problem?

anonymous · Answer

Unnecessary but you can...
a=-10
dv/dt=-10
Multiply and divide by dx, dx/dt=v
so v dv/dx=-10
v dv = -10dx
v^2/2=-10h putting limits for v as 96 to 0 you get the answer.

anonymous · Answer

ITs basically the same all newtons laws are derived from these equations only.

anonymous · Answer

we must convert feet to meter?

anonymous · Answer

i dont understand how 10 dx became h?

anonymous · Answer

i know that....

anonymous · Answer

I integrated both sides limits are v to 0 for dv and 0 to h for dx so you get h.