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

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

Physics
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takeoff speed of the athlete
|dw:1433534380042:dw| do you recall the equations for x(t) and y(t) ??

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i am not sure of which equations to use
is it x= x= x0+Vx0t-1/2gt^2?
here 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...
yes that was for y not x
is x0 the 7.68?
i am confused about which numbers are what variables
x0 and y0 are the coordinates of the starting point, in this problem we can set both of them to 0. |dw:1433535031686:dw|
and xt?
x(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
so i have to figure out the y component first?
you 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...
our unknows here are v0 and t, so we must solve the system for both of them.
so the equation would be Vo(t)-1/2gt^2?
1/2gt-Vsin(33.4)??
t=2Vsin(33.4)/g??
remember 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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