A red ball is thrown down with an initial speed of 1.1 m/s from a height of 29.0

Question

A red ball is thrown down with an initial speed of 1.1 m/s from a height of 29.0 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 25.2 m/s, from a height of 1 meter above the ground. g= 9.81 m/s2. -->How long after the red ball is thrown are the two balls in the air at the same height?

Explanation / Answer

we have the formula s=si + ut + 1/2 at^2 (s=displacement, si=initial displacement, t=time, a=accelration now s(r)=s(b) therefore si(r)+u(r)(t+0.6) +1/2 a (t+0.6)^2= si(b) + u(b)t + 1/2 a t^2 => 29-1.1x(t+0.6) - 1/2g(t+0.6)^2 = 0.7 + 25.5t -1/2gt^2 =>29-0.7 -0.66-1.1t-25.5t -1/2g(o.6)^2 - g(0.6t)=0 =>27.64-4.905x.36-26.6t-9.81x.6t=0 =>27.64-1.7658 - 26.6t-5.886t=0 =>25.8742=32.486t =>t=25.8742/32.486t =>t=0.7965 seconds so 0.6+0.7965 seconds after red ball is thrown , the two ball are at the same height answer 1.396 seconds

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