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anonymous
 one year ago
An athlete performing a long jump leaves the ground at a 33.4∘ angle and lands 7.68m away.
anonymous
 one year ago
An athlete performing a long jump leaves the ground at a 33.4∘ angle and lands 7.68m away.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you want to ask? you have provided the data.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0takeoff speed of the athlete

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433534380042:dw do you recall the equations for x(t) and y(t) ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am not sure of which equations to use

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it x= x= x0+Vx0t1/2gt^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here they are: \[x(t)=x_0 + v_{0x}t\] \[y(t)=y_0 + v_{0y}t  \frac{1}{2}gt^2\] the conditions are in the picture i made... try to put all that togheter...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that was for y not x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am confused about which numbers are what variables

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x0 and y0 are the coordinates of the starting point, in this problem we can set both of them to 0. dw:1433535031686:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x(t) is the value x takes at time t. we know that for some value of time x(t)=7.68m, we dont know the value of t, but at the same time y(t) must be 0, when the atlete reaches the ground

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i have to figure out the y component first?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you would have to find the value of t for which y(t)=0, you can do so setting: \[y(t)=0\] but using the whole equation for y(t), and finding t from there...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0our unknows here are v0 and t, so we must solve the system for both of them.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the equation would be Vo(t)1/2gt^2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0remember t was squared... so: \[v_0 sin(33.4º)t\frac{1}{2}gt^2 = 0\] is the first equation and \[v_0cos(33.4º)t=7.68\] is the other one, in order to find v0 we need to solve the system... first can be rewritten to: \[v_0sin(33.4º)t=0.5\times9.81\times t^2\] so if we take the quotient between them ( this is amethod to solve systems ) we get: \[\frac{v_0sin(33.4)t}{v_0cos(33.4)t} = \frac{0.5\times 9.81 \times t^2}{7.68}\] so v0 and t cancel out in left hand side and you can get t from there, then you use that value of t in any of the original equations to find v0.
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