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>
plug in your numbers where 4=x1, 3=y1, 6=x2, and 7=y2
which equals 2