Solving Ball Throw Problem: Velocity & Distance for Boy & Dog

[k2var2002]

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$$\theta$$*t-1/2gt^2

1/2gt/yo= voy

R= Voy*t=Vosin$$\theta$$*t

t= $$\sqrt{}$$(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

 

[tiny-tim]

Welcome to PF!

Hi Kyle! Welcome to PF!

(have a square-root: √ and try using the X² tag just above the Reply box )

2. y=yo+voy*t-1/2gt^2

How did you get t= √(2y/g)^2 from that?

 

[k2var2002]

t= √(2y/g)^2 is just a derivation of some sort that my professor had given us to figure time in 2d motion.

 

[k2var2002]

Would t=Vo/-a work from the base equation Vy=Vo+at?

 

[tiny-tim]

t= √(2y/g)^2 is just a derivation of some sort that my professor had given us to figure time in 2d motion.

That only works if v₀ = 0 … just look at the original equation.

Stop looking for shortcuts!