Asked by Anonymous
An object is thrown at 30 m/s and at a 40 degree angle. What is the range of the object, given that the ground is horizontal and flat?
I'm not sure whether to use (vi^2sin2x)/g or x = Vh*t.
I'm not sure whether to use (vi^2sin2x)/g or x = Vh*t.
Answers
Answered by
Damon
Vi = 30 sin 40
u = 30 cos 40
h = 0 + Vi t - 4.9 t^2
0 = 30 sin 40 t - 4.9 t^2
0 = t (30 sin 40 - 4.9 t)
so
t = 30 sin 40/4.9 for downrange time
range = r = u t
r = 30 cos 40 * 30 sin 40 /4.9
= 30^2 sin 80/9.8
so the first one :)
u = 30 cos 40
h = 0 + Vi t - 4.9 t^2
0 = 30 sin 40 t - 4.9 t^2
0 = t (30 sin 40 - 4.9 t)
so
t = 30 sin 40/4.9 for downrange time
range = r = u t
r = 30 cos 40 * 30 sin 40 /4.9
= 30^2 sin 80/9.8
so the first one :)
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