A shot putter releases the shot some distance above the level ground with a velo

Question

A shot putter releases the shot some distance above the level ground with a velocity of 10.2 m/s, 53.0 degree above the horizontal. The shot hits the ground 2.16 s later. You can ignore air resistance. (Assume the horizontal direction of the shot and upward are positive.) (a) What are the components of the shot's acceleration while in fight? m/s^2 (horizontal component) m/s^2 (vertical component) (b) What are the components of the shot's velocity at the beginning of its trajectory? m/s (horizontal component) m/s (vertical component) What are the component is of the shot's velocity at the end of its m/s (horizontal component) m/s (vertical component) (c) How far did she throw the shot horizontally? m (d) the expression for R (as given below) not give the correct answer for part (c)? R = v_0^2 sin 2 sigma_0/g The initial and final heights are different. It is not a projectile motion. There is air drag. (e) How high was the shot above the ground when she released it? m

Explanation / Answer

a)horizontal acceleration is 0
vertical acceleration is -9.81m/s^2 due to gravity

b)
10.2*sin(53) = 8.146 m/s vertical
10.2*cos(53) = 6.14 m/s horizontal

8.146 - (9.81)(2.16) = -13.04 m/s vertical
6.14 m/s = horizontal

c) d=(6.14)(2.16) = 13.26 m

e) s = ut + 0.5a(t^2) . = 8.146*2.16 + 0.5*(-9.81)*2.16^2 = -5.289 m

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