## idku one year ago One question.

1. idku

 Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.  |dw:1441299455460:dw|

2. idku

theta = 44° and I need the distance between the two people

3. geerky42

Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance.

4. idku

I have found so far: 1) Horizontal Component of the Velocity: $$\$0.8em]$$ $$V_x~~=~~V_0\times \cos(\theta)~~=~~11\times \cos(44)~~=~~11\cos(44)$$ 2) Vertical Component of the Velocity: $$\\[0.8em]$$ $$V_y~~=~~V_0\times \sin(\theta)~~=11\times \sin(44)~~=~~11\sin(44)$$ 3) Maximum Height: $$\\[0.8em]$$ $$V_y~~=~~V_{0y}-gt$$ $$0~~=~~11\sin(44)-9.81t$$ so max height ocucrs at $$t=11\sin(44)/9.81$$ and now we plug that into vertical displacement: $$y=V_{0y}\times t+y_0$$ $$y=11\sin(44)\times \left( 11\sin(44)~/~9.81\right)+1.5 \approx 4.5 ~~({\rm meters})$$ 5. idku What formula would I use to find the time that takes the ball to fly from one person to another? 6. idku Disregard the previous long 3 part reply, that is more for myself. 7. geerky42 Because initial height and final height are same, we can just double the time it takes for ball to reach maximum height. 8. IrishBoy123 yeah, you don't need to know how high it went, just how long it was in the air for 9. idku Can we go by the first thing you proposed please? $$\color{blue}{\text{Originally Posted by}}$$ geerky42 Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance. $$\color{blue}{\text{End of Quote}}$$ 10. idku Which formula am I starting to use, can you please tell me? 11. geerky42 You figured out the time it takes for ball to reach max, so just double it (2t), then we got time we need. 12. idku ok, so we need 22sin(44) / 9.81 seconds, (roughly 4.5 × 2 = 9) seconds. 13. idku ok, and then how would I use the time between them to find the distance between them> 14. geerky42 Just focus on horizontal component. We know that horizontal velocity doesn't change, so formula is simply $$d = vt$$ we have time and horizontal velocity, so we can multiply them to get distance. 15. geerky42 Makes sense, right? 16. idku oh, so d=11cos(44)×(11sin(44)÷9.81)= (11/19.62) sin(88) 17. idku 11cos(44) is v 11sin(44) / 9.81 is time 18. geerky42 actually time is 22sin(44)/9.81 remember we need time it takes for ball to go from Julie to Sarah. 19. idku Ok, 12.32 is wolfram's approximation. How does it exactly work.....? d=vt Only gives us 1/2 of the distance for some reason? 20. idku or rather full time is 2t. 21. idku so t is 2t:)(()()()() 22. geerky42 Yeah 23. idku Ok, I am very curious, because on my test it would probably not be this way. Can you expain how would you go about doing your original approach that you proposed?  Plan is to figure out the time it takes for ball to go from Juile to Sarah. For that, we focus on vertical component. After that, we can focus on horizontal component and figure out the distance.  24. geerky42 More of less same as what we just did. Horizontal component is very simple $$d = vt$$, however we only know the value of $$v$$. So I'd just calculate full time with given vertical velocity by using $$V_y = V_{o_y}−gt$$ Initial velocity and final velocity should be opposite, right? So we have \[-V_{o_y} = V_{o_y}-gt\quad\Longrightarrow\quad\dfrac{2V_{o_y}}{g}=t$ Now we got $$t$$, so we then calculate distance.

25. idku

Oh Vy=Voy - gt is what we would have used. Tnx

26. geerky42

No problem