Asked by Benjamin
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?
marshmallow at an angle of 30 degrees. How hard did she throw the marshmallow?
Answers
Answered by
Damon
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)
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)
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