help please only have little time left just tiny part of question!!!
The horizontal velocity is 10.5 cos(24.1). The vertical velocity is 10.5sin(24.1)-0.36*9.81 (you have calculated both of these to answer part A, 9.61 m/s). Then horiz.vel./vert.vel. is tangent(theta) where theta is angle required.
soo i got 65.9 does that sound correct?
oops its nots
Not exactly. The answer is way smaller. Remember there is a subtraction in the vertical velocity.
oops!
Correction should read: \( \large \frac{vert. velocity}{hor.velocity} = tan(\theta) \)
so do i use arctangent after i divide ver vel/hor vel?
Yep!
so arctangent 12.62= 85.47
but thats not smaller than my other one hmm
What did you get for vert. vel. and hor. vel.?
You have probably inverted horiz. and vert.
horiz= 9.58?
9.58 is right.
vert= .7559
That is correct too!
Remember I made a mistake in my previous post, and later corrected?
i got 4.51?
That is correct! Good job!
its positive right not neg?
It's positive because both vert and horiz. vel. are positive.
man, it said it was wrong?
any ideas on why? i only have one chance left :/ PS-thanks for all your help your awesome
Sorry, I got the wrong question. Time was 0.72 seconds. So you can repeat the same calculations replacing 0.36 sec. by 0.72 sec, but the angle will be negative because the vertical velocity will be negative.
The reason it was wrong was 0.36 second apply to the first two questions. The last two questions were for 0.72 second after the ball was kicked. I did not read the third question. Sorry. I get -16.15 deg. (note: it's negative).
-16.12 sound correct?
oh hmm
We're pretty close! :)
do you have a few seconds to look at my last question/???
I'll check.
The "vector A" question?
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