At 2/3 of its maximum height, it has 1/3 of its initial kinetic energy left. That means it has 1/sqrt3 = 0.5773 of its initial velocity. The initial velocity must have been
Vo = 56/0.5773 = 96.99 m/s.
Max height is Vo^2/(2*g)
= 480 m
An object is thrown vertically upward such
that it has a speed of 56 m/s when it reaches
two thirds of its maximum height above the
launch point.
Find the maximum height h. The acceleration of gravity is 9.8 m/s
2
.
Answer in units of m
2 answers
v.=56m/s
v=0
g=-10m/s2
v2=V.2+2gyx
0=3136-2.9.8.x
19.6x=3136m
x=160m
1/3x=160m
x=3.160=480m
v=0
g=-10m/s2
v2=V.2+2gyx
0=3136-2.9.8.x
19.6x=3136m
x=160m
1/3x=160m
x=3.160=480m
1 answer