   Chapter 12.3, Problem 41E

Chapter
Section
Textbook Problem

# Shot-Put Throw The path of a shot thrown at an angle θ is r ( t ) = ( v 0 cos θ ) t i + ⌈ h + ( v 0 sin θ ) t − 1 2 g t 2 ⌉ j where v 0 is the initial speed, h is the initial height, t is the time in seconds, and g is the acceleration due to gravity. Verify that the shot will remain in the air for a total of t = v 0 sin θ + v 0 2 sin 2 θ + 2 g h g secondsand will travel a horizontal distance of v 0 2 cos θ g ( sin θ + sin 2 θ + 2 g h v 0 2 ) feet.

To determine

To Prove: The shot will remain in the air for a total of t=v0sinθ+v02sin2θ+2ghg seconds and will travel a horizontal distance of v02cosθg(sinθ+sin2θ+2ghv02) feet if the path of a shot thrown at an angle θ is r(t)=(v0vosθ)ti+[h+(v0sinθ)t12gt2]j.

Explanation

Given:

Time is t=v0sinθ+v02sin2θ+2ghg seconds.

Horizontal distance is v02cosθg(sinθ+sin2θ+2ghv02) feet.

Path of a shot is r(t)=(v0vosθ)ti+[h+(v0sinθ)t12gt2]j.

Formula used:

x=b±b24ac2a.

Proof:

Consider the position vector of the projectile,

r(t)=(v0vosθ)ti+[h+(v0sinθ)t12gt2]j

To find the range setting substitute y=0,

y=h+(v0sinθ)t12gt2=0

(12g)t2(v0sinθ)th=0t=v0sinθ

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