- #1
k2var2002
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1. A boy 11.0m above the ground in a tree throws a ball for his dog, who is standing
right below the tree and starts running the instant the ball is thrown.
If the boy throws the ball upward at 40.0° above the horizontal, at 7.50m/s .
What is velocity (m/s)? and distance (m)?
2. y=yo+voy*t-1/2gt^2
y= yo+sin[tex]\theta[/tex]*t-1/2gt^2
1/2gt/yo= voy
R= Voy*t=Vosin[tex]\theta[/tex]*t
t= [tex]\sqrt{}[/tex](2y/g)^2
3. I plugged the known variables into the equations i listed above. First for time, I got t=1.5s. Then I used the angle and time found to find distance, R=(7.5)(sin40)=4.82 m. Which doesn't make sense to me. Any help is appreciated. Thanks!
-Kyle
right below the tree and starts running the instant the ball is thrown.
If the boy throws the ball upward at 40.0° above the horizontal, at 7.50m/s .
What is velocity (m/s)? and distance (m)?
2. y=yo+voy*t-1/2gt^2
y= yo+sin[tex]\theta[/tex]*t-1/2gt^2
1/2gt/yo= voy
R= Voy*t=Vosin[tex]\theta[/tex]*t
t= [tex]\sqrt{}[/tex](2y/g)^2
3. I plugged the known variables into the equations i listed above. First for time, I got t=1.5s. Then I used the angle and time found to find distance, R=(7.5)(sin40)=4.82 m. Which doesn't make sense to me. Any help is appreciated. Thanks!
-Kyle