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Anna [14]
3 years ago
10

I NEED HELP PLEASE, THANKS! :)

Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0

Answer:

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

Step-by-step explanation:

If we are given that an object is thrown with an initial velocity of say, v1 m / s at a height of h meters, at an angle of theta ( θ ), these parametric equations would be in the following format -

x = ( 30 cos 20° )( time ),

y = - 4.9t^2 + ( 30 cos 20° )( time ) + 2

To determine " ( 30 cos 20° )( time ) " you would do the following calculations -

( x = 30 * 0.93... = ( About ) 28.01t

This represents our horizontal distance, respectively the vertical distance should be the following -

y = 30 * 0.34 - 4.9t^2,

( y = ( About ) 10.26t - 4.9t^2 + 2

In other words, our solution should be,

x = 28.01t,

y = 10.26t - 4.9t^2 + 2

<u><em>These are are parametric equations</em></u>

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