Question

A rock is thrown straight upward with an initial velocity of 9.6 m/s in a location where the acceleration due to gravity ahs a magnitude of 9.81 m/s^2. To what height does it rise?

please walk me through this problem:)

Answer 1

When it stops rising, its speed is zero
V = Vi - 9.81 t
0 = 9.6 - 9.81 t
t = 9.6/9.81
then use
h = ho + Vi t - (9.81/2)t^2
h = 0 + 9.6 (9.6/9.81) - (9.81/2)(9.6/9.81)^2

Answer 2

There is also a shortcut way to do this using "energy conservation" that will give the same answer. At the maximum height H, the potential energy increase M g H will equal the initial kinetic energy, (1/2) M Vo^2)

Therefore H = Vo^2/(2g)