Asked by Ellie
Jordan throws a golfball straight up from the top of a building 50 meters high. The initial speed of the golfball is 20 m/s and it just misses the edge of the roof when it eventually falls. Neglecting air resistance, find the velocity of the golfball just at the instant it hits the ground.
Answers
Answered by
Henry
h = ho + (V^2-(Vo^2))/2g
h = 50 + (0-(20^2))/-19.6 = 70.4 m. Above gnd.
V^2 = Vo^2 + 2g*h = 0 + 19.6*70.4 = 1380
V = 37.1 m/s.
h = 50 + (0-(20^2))/-19.6 = 70.4 m. Above gnd.
V^2 = Vo^2 + 2g*h = 0 + 19.6*70.4 = 1380
V = 37.1 m/s.
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