Rock #1:
Vo1 = 7.70m/s[10o]. Downward.
Yo1 = 7.70*sin10 = 1.34 m/s.
h1 = Yo1*t + 0.5g*t^2
1.34*3.7 + 4.9*3.7^2 = 72.0 m Below top of the tower.
Rock #2:
Vo2 = 7.70m/s[50o], Up.
Yo2 = 7.70*sin50 = 5.90 m/s
h2 = Yo2*t + 0.5gt^2
h2 = 5.90*3.7 - 4.9*3.7^2 = -45.3 m. =
45.3 m. Below the top of the tower.
h1-h2 = 72 - 45.3 = 26.7 m. From each other.
Two rocks are thrown simultaneously from the top of a very tall tower with identical speeds of v = 7.70 m/s, but in two different directions.One of them is thrown with an angle of 10.0 degrees below the horizontal, the other one at an angle of 50.0 degrees above the horizontal. How far will the rocks be from each other after 3.70 s? (Neglect air resistance and assume that the rocks will not hit the ground.)
I don't even know where to begin on this one. Someone please help!
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