A stone is thrown vertically upward at a speed of 49.80 m/s at time t=0. A second stone is thrown upward with the same speed 2.670 seconds later. At what time are the two stones at the same height?(adsbygoogle = window.adsbygoogle || []).push({});

i found that to be 6.411s

at what hight do the two stones pass eachother?

ok, so that hight will be at the time that they are at the same hight which is t= 6.411s

so i used the equation deltax=Vot+1/2at^2 and i got 1023 m but that isn't right.... can someone help me out please

also....

what is the downward speed of the first stone as they pas eachother?

so that time again would be 6.411

and i used the equation v=vo+at .....so v=49.80-9.8(6.411) which is got to be -13.32 m/s but that isn't right either

please help me

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# Basic physics prob.

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