Vo = 6.2m/s[10o]
Xo = 6.2*cos10 = 6.106 m/s.
Yo = 6.2*sin10 = 1.077 m/s.
1. h = ho + (Y^2-Yo^2)/2g
h = 11 + (0-(1.077)^2)/-19.6 = 11.06 m.
Above gnd.
2. h = 0.5g*t^2 = 11.06 m.
4.9t^2 = 11.06
t^2 = 2.257
t = 1.50 s.
3. d = Xo * t = 6.106m/s * 1.50s=9.17 m.
4. Y^2=Yo^2 + 2g*h= 0 + 19.6*11.06=216.8
Y = 14.72 m/s.
Tan A = Y/Xo = 14.72/6.106 = 2.41129
A = 67.5o
A playful college student tosses a water balloon out of his dorm window at a speed of 6.2 m/s at an angle of 10° above the horizontal. The dorm window is 11 m above ground level.
1)What is the maximum height above the ground of the balloon?
2) At what time does the balloon strike the ground?
3) At what distance from the dorm does the balloon hit the ground?
4) With what angle does the balloon strike the ground?
2 answers
the answer is wrongg ;(