# How to work out the horizontal and vertical velocity if I am not given the time takenWatch

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#1
An object was released from a cable car travelling at a speed of 4.6ms-1 in a direction of 40 degrees above the horizontal.

Calculate the horizontal and vertical components of velocity of the object at the instant it was released.
How do I work this out. If I was given the time it took I could use y = U t sin(0) - gt and v = U*sin(0) - gt but I am not given it.
0
3 weeks ago
#2
You don’t need time
It’s just trigonometry with angle 40 and hypotenuse of 4.6ms^-1
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