Simple question. But trick involved. A ball is thrown horizontally from the top of a 10 meter high builiding. It strikes the ground at an angle of 30 degree. with what speed was it thrown?

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Simple question. But trick involved. A ball is thrown horizontally from the top of a 10 meter high builiding. It strikes the ground at an angle of 30 degree. with what speed was it thrown?

Physics
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was the angle with respect to ground ?

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Other answers:

Is it 20√6m/s
thrown horizontally means angle was 90 degree |dw:1433846133143:dw| assuming no drag force , so the horizontal velocity is constant call it V0 and the vertical velocity is Vy =zero so the final vertical velocity is \[V _{T}^{2}-V_{y}^{2} = 2 g h\] thus you have vy |dw:1433846575866:dw| so \[\tan 30 = \frac{ V _{0} }{ V _{y} }\rightarrow V _{0}= V _{y}* \tan30\] just plug in numbers
uh..oh I mixed Vt with Vy. hope you get the main point
hey, using the equation you gave me, the answer comes out to 8.1. But the actual answer is 24.
What i did was take the time it takes get to the bottom. Then i found out the horizontal distance using trig. Then i divided distance by time. Why is that wrong?
@curry, what time did you get for the ball to fall that vertical distance of 10 meters? I got 1.43 seconds.
|dw:1433885146576:dw| @Catch.me
Uh Oh gotta go now, but will return to see how this turned out lol
Yes, i got 1.43 seconds for time also. And then for the horizontal distance i got 10rad3. So i divided this value by 1.43 to get the horizontal speed. But that's not giving me the right answer.
|dw:1433887281746:dw| This is the diagram given. With the curved path being the path of the ball.
Wait I think catch.me is correct actually. The time it takes to get to the bottom is 1.42 seconds. So the Vy is 9.8*1.42. tan(30) = Vy/Vx. Giving us a value of 24 m/s. I understand why this way works. But I still don't understand why the other way doens't work.
\(10 \sqrt{3}\)?
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@curry medal to @Catch.me would be fair. at time of writing, none.
@radar yeah i should be 60 @Curry how did you calculate horizontal distance?
|dw:1434062799673:dw| @Curry I used the law of sines. Sin 60/x = Sin 30/10. I'm not sure if that would of given me the horizontal distance, but that is what I used.
The curve is parabola not straight line( angle is changing) vt= 14 m/s v0= 14 root 3 time = delat v/g=10/7 x= 20 root 3

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