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idku
 one year ago
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idku
 one year ago
One question.

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idku
 one year ago
Best ResponseYou've already chosen the best response.0``` 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

idku
 one year ago
Best ResponseYou've already chosen the best response.0theta = 44° and I need the distance between the two people

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4Plan 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0I 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: \(y=V_{0y}\times t+y_0\) \(y=11\sin(44)\times \left( 11\sin(44)~/~9.81\right)+1.5 \approx 4.5 ~~({\rm meters})\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0What formula would I use to find the time that takes the ball to fly from one person to another?

idku
 one year ago
Best ResponseYou've already chosen the best response.0Disregard the previous long 3 part reply, that is more for myself.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4Because initial height and final height are same, we can just double the time it takes for ball to reach maximum height.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0yeah, you don't need to know how high it went, just how long it was in the air for

idku
 one year ago
Best ResponseYou've already chosen the best response.0Can 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. \(\color{blue}{\text{End of Quote}}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0Which formula am I starting to use, can you please tell me?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4You figured out the time it takes for ball to reach max, so just double it (2t), then we got time we need.

idku
 one year ago
Best ResponseYou've already chosen the best response.0ok, so we need `22sin(44) / 9.81` seconds, (roughly 4.5 × 2 = 9) seconds.

idku
 one year ago
Best ResponseYou've already chosen the best response.0ok, and then how would I use the time between them to find the distance between them>

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4Just 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0oh, so d=11cos(44)×(11sin(44)÷9.81)= (11/19.62) sin(88)

idku
 one year ago
Best ResponseYou've already chosen the best response.011cos(44) is v 11sin(44) / 9.81 is time

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4actually time is 22sin(44)/9.81 remember we need time it takes for ball to go from Julie to Sarah.

idku
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 12.32 is wolfram's approximation. How does it exactly work.....? d=vt Only gives us 1/2 of the distance for some reason?

idku
 one year ago
Best ResponseYou've already chosen the best response.0or rather full time is 2t.

idku
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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. ```

geerky42
 one year ago
Best ResponseYou've already chosen the best response.4More 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.

idku
 one year ago
Best ResponseYou've already chosen the best response.0Oh Vy=Voy  gt is what we would have used. Tnx
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