Question

In the Olympic shotput event, an athlete throws the shot with an initial speed of 11.0 m/s at a 44.0° angle from the horizontal. The shot leaves her hand at a height of 1.80 m above the ground

How far does the shot travel?

Express your answer with the appropriate units.

Answer 1

Sorin Voyî ve Filmes hy h=1.8om | - During the flight time t, total vertical displacement by = Voyt + hay ta. Where ay = o-h=th=-1.80m. Voy = vertical speed linitial) =(11 m/s) sinnu 1 = 7.64 m/s ay =-9.80 m/s2 t = ? Now, -1.60=(7.64m/s) t+ – 2 1-9.80 m/s2) Z2 4.90t3-71647-1.80 = 0 By sowing the equation t=10775 → Hosizontal range (R) = Vont a 1 Where von initial horizontal speed =(11 m/s) c'est = 7.9, mis S; R =(1.91m/s) (1.775) R = 14m] ---- Answer