A stone is thrown vertically upward at a speed of 15.00 m/s at time t=0. A secon

Question

A stone is thrown vertically upward at a speed of 15.00 m/s at time t=0. A second stone is thrown upward with the same speed 0.450 seconds later. At what time are the two stones at the same height?
ANS=1.754 s But I will like to know At what height do the two stones pass each other?

Explanation / Answer

the height of the first stone: h = v0 t - (1/2)g t^2 v0 = 15/s for second stone, we have: h1 = v0 (t-0.45s) -(1/2)g(t-0.45s)^2 when they have the same height, we have: h=h1, => v0 t - (1/2)g t^2 = v0(t-0.45) - (1/2)g(t-0.45)^2 15t-0.5*9.8*t^2 = 15t-15*0.45-0.5*9.8*(t^2+0.45^2-2t*0.45) 15t-0.5*9.8*t^2 = 15t-15*0.45-0.5*9.8*t^2-0.5*9.8*0.45^2+2*t*0.45*0.5*9.8 -0.5*9.8*t^2=-15*0.45-0.5*9.8*t^2-0.5*9.8*0.45^2+2*t*0.45*0.5*9.8 -4.9t^2=-6.75-4.9t^2-0.99+4.41t 4.41t=6.75+0.99 t=1.754 s height do the two stones pass each other h = v0 t - (1/2)g t2 =15m/s*1.754s - (1/2)*g*(1.754s)2 = =(15*1.754)-(0.5*9.8*1.754*1.754) = 11.23m

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