An soccer ball is launched horizontally off a cliff that is 12 meters tall with in initial velocity of 5 meters per second. The ball experiences a resistant force where the ball decelerates by ax=3 m/s2 in the x direction and ay=3.8 m/s2 in the y direction. How long does the object fall for?
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First pick a coordinate system. Let x positive be 'right' y positive be up the ball accelerates down because of gravity, and accelerates
So that's what's happening. Always start out with a diagram for physics problems.
Let's also write down all our givens: \[\sf V_{\large i_x} = 5 \frac{m}{s}\]\[\sf a_x = 3\frac{m}{s^2}\]\[\sf a_y =3.8\frac{m}{s^2}\]
I believe a_x = -3 m/s^2, since it is opposite to the direction of motion
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ay = -9.8 m/s^2 + 3.8 m/s^2
I set downwards to being positive :P
ok :)
so you can solve this using calculus, taking antiderivatives
ax = -3 ay = -9.8 + 3.8 = -6 vx = -3*t + c1 vy = -6t + c2
note that i chose +x to the right, +y up
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