anonymous
  • anonymous
. A golf ball is hit upward at an angle θ from the horizontal and speed v = 40 m/s. It reaches a maximum height of 10 m as illustrated above. Assume the ballistic trajectory starts at ground level and ignore air resistance. Assume the shot is made on a wide, level field. What is the initial angle θ ? A) θ = 20.5° B) θ= 46.4° C) θ = 57.3° D) θ = 70.2° E) θ = 81.1° 31. How long is the ball in the air? A) 1.15 s B) 2.86 s C) 3.12 s D) 3.37 s E) 3.59 s
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
souvik
  • souvik
first divide the velocity of the ball in two components..vertically and horizontally... there is a downward acceleration due to the pull of gravity.. and u know the maximum height of the ball..its 10 m use.. \[v ^{2}_{y}=u ^{2}_{y}-2gH\] where \(v_{y}\)=final vertical velocity \(u_{y}\)=initial vertical velocity H=maximum height u can get the angle \(\theta\) from here...
souvik
  • souvik
and for the second question..use this...\[v _{y}=u _{y}-g T/2\] where T is the time of flight

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souvik
  • souvik
in the final position \(v_{y}\)=0 and at initial position \(u_{y}\)=\(u sin\theta\)
anonymous
  • anonymous
can you explain how to get thetha from the first equation?? @souvik
souvik
  • souvik
here the vertical component of the velocity\(u_{y}\)=\(u sin\theta\)

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