Question

If you throw a 0.3kg ball straight up with an initial speed of 39m/s, how fast will it be moving when it’s 21m above the release point?

Answer 1

The question relies on conservation of energy

2 Energy equations:

Kinetic Energy = 1/2mv^2

Gravitational potential energy = mgh

Initial energy = final energy

1/2m(v0)^2 = mgh + 1/2m(vf)^2

Divide mass out to save time

1/2(v0)^2 = gh + 1/2(vf)^2

PLug in what we know:

v0 = 39 m/s

h = 21 meters

1/2(39)^2 = 9.8(21) + 1/2(vf)^2

1/2(39)^2 - 9.8(21) = 1/2(vf)^2

554.7 = 1/2(vf)^2

1109.4 = (vf)^2

vf = 33.3076 m/s