Question

An object is thrown vertically upward and has a speed of 10 m/s when it reaches three fifths of its maximum height above the launch point. Determine its maximum height.

Answer 1

i used the 2ad = V(inital)^2 - V^2 formula
H = (1/5), a = -9.8, V^2 = 10m/s since intial V is 0

so i had 2(-9.8)d = 0^2 - 10^2

-19.6d = -100 => d= -100/-19.6 => d=5.10m after this i multiplied it by the reciprocal of 1/5 and got 25.5m i'm stuck on where i went wrong?

Answer 2

We know
\[v^2-u^2=2as\]
v= final velocity at max height is 0
u = u ( initial velocity)
s= h (max height )
a=-9.8 m/s^2
so we get
\[-u^2=-2*9.8*h.............equation 1\]
now at s= 3/5 h
v= 10 m/s
u= u m/s
so from this
we get
\[10^2-u^2=-2*9.8*\frac{3}{5}*h..........equation 2\]
subtract equation 1 from 2
we get
\[10^2=-2*9.8*\frac{3}{5}h+2*9.8*h\]
on simplifying we get
\[100=2*9.8*\frac{1}{5}*h\]
so
\[h=\frac{100*5}{2*9.8}\]
\[h = 25.51 m\]
so max height is 25.51 meters

Answer 3

you are correct galactic