Question

A girl throws a marshmallow that lands in her friend’s mouth 2 m away. The girl threw the
marshmallow at an angle of 30 degrees. How hard did she throw the marshmallow?

Answer 1

well, I guess I assume the mouth and the throwing hand are at the same height, call it h = 0
S is the speed we want

u = S cos 30
Vi = S sin 30
v =
h = 0 + Vi t - 4.9 t^2
when is h = 0 (same height as at start)
4.9 t^2 -Vi t = 0
(4.9 t -Vi)t = 0
so t = 0 (that was at the start
or t = Vi/4.9
so
Vi = 4.9 t = S sin 30
t = S sin 30/4.9

now the horizontal problem
2 meters = u t
2 = S cos 30 * t
t = 2/(S cos 30) = S sin 30/4.9
9.8 = S^2 sin 30 cos 30
S^2 = 9.8 / (sin 30 * cos 30)