Asked by walter
You throw a snowball at an angle of 34.0° below the horizontal from the roof of a building that is 74.0 m above flat ground. The snowball hits the ground a distance of 44.0 m from the base of the building. Assume free fall with up as positive, and use g = 9.80 m/s2.
how long the projectile remain in the air?
what is the initial speed of the snowball?
how long the projectile remain in the air?
what is the initial speed of the snowball?
Answers
Answered by
Henry
Dx = Vo^2*sin(2A)/g = 44. m.
Vo^2*sin(68)/9.8 = 44.
Vo^2*0.0946 = 44.
Vo^2 = 465.1.
Vo = 21.6 m/s. at 34o below the hor.
Xo = Vo*Cos A = 21.6*Cos34 = 17.9 m/s.
Dx = 17.9*T = 44.
T = 2.46 s. = Time in air.
Vo^2*sin(68)/9.8 = 44.
Vo^2*0.0946 = 44.
Vo^2 = 465.1.
Vo = 21.6 m/s. at 34o below the hor.
Xo = Vo*Cos A = 21.6*Cos34 = 17.9 m/s.
Dx = 17.9*T = 44.
T = 2.46 s. = Time in air.
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