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?

Question

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?

jhannybean · Answer

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perl · Answer

First pick a coordinate system. Let x positive be 'right' y positive be up 
the ball accelerates down because of gravity, and accelerates

jhannybean · Answer

So that's what's happening. Always start out with a diagram for physics problems.

jhannybean · Answer

Let's also write down all our givens: $$\sf V_{\large i_x} = 5 \frac{m}{s}$$$$\sf a_x = 3\frac{m}{s^2}$$$$\sf a_y =3.8\frac{m}{s^2}$$

perl · Answer

I believe a_x = -3 m/s^2, since it is opposite to the direction of motion

jhannybean · Answer

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perl · Answer

ay = -9.8 m/s^2 + 3.8 m/s^2

jhannybean · Answer

I set downwards to being positive :P

perl · Answer

ok :)

perl · Answer

so you can solve this using calculus, taking antiderivatives

perl · Answer

ax = -3 
ay = -9.8 + 3.8 = -6

vx = -3*t + c1
vy = -6t + c2

perl · Answer

note that i chose +x to the right, +y up