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an object is thrown up into the air going 9 m/s. How fast is it going 2 seconds later?

i dont get how to do it

Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
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This is the problem on uniformly accelerated motion. In this case, the object is thrown up, but the gravity pulls it down, with acceleration 9.8 m/s2. Because the acceleration is directed opposite of the velocity, velocity will decrease. Since gravitational acceleration is usually assumed to be constant, this is uniformly decelerated motion. In this case velocity changes by the same amount each second, namely by 9.8 m/s every second. So,

vf=vi-gt, where vi and vf are initial and final velocities, g=9.8 m/s2, t is time elapsed. vi=9 m/s. Thus,

vf=9-9.8*2=9-19.6=-10.6 m/s. Negative sign means that the velocity is now directed opposite of the initial direction, which was upward direction. Therefore, it is now directed downward and its magnitude is 10.6 m/s.

Notice, that if the object is thrown from the ground level, its speed at the impact moment must be the same as the one it was tossed up with, due to symmetry considerations. However, its speed is greater than initial speed. So there may be a trick in this question: if the object is assumed to be tossed up from the ground level, in 2 s it has already hit the ground and its velocity is zero. Or we can add another dimension to this and assume that the object bounces back from the ground after the impact. In this case, we need to find the time it hit the ground at. Since its speed must be 9 m/s at the impact, pointing downward, its final velocity is -9 m/s, therefore the time it took to hit the ground again is:

t0=(vi-vf)/g=(9-(-9))/9.8=18/9.8=90/49≈1.84 s.

Assuming perfectly elastic bounce and negligible time of impact, we obtain that the object bounced back at 1.84 s and its time of flight upward for the second time is tf=2-90/49=8/49≈0.16 s. Then again solve the first equation for vf, this time with tf to obtain:

vf=9-9.8*8/49=9-1.6=7.4 m/s.

We could keep going and lift up the assumption of a perfectly elastic bounce or introduce the finite impact time. :-) :D I hope you do not take all this too seriously, my purpose is to illustrate that the problem is underformulated and some important information is missing from its statement. I would assume that the question was to just formally find the final velocity and do not bother about all those details I pointed out in my funny addition.