Asked by girl
A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.3 s later it is rising at a speed of 18 m/s. Assuming air resistance has no effect on the rock, calculate its speed at (a) at launch and (b) 5.4 s after the launch.
Answers
Answered by
Elena
v₁=v₀-gt₁
v₀=v₁+ gt₁=18+9.8•5.4= 40.54 m/s
At the top point v=0
0 =v₀-gt
t = v₀/g=40.54/9.8 =4.14 s.
t₃=t₂- t=5.4 – 4.14 = 1.26 s
v₂=gt₃ = 9.8•1.26 = 12.38 m/s (directed dowmward)
v₀=v₁+ gt₁=18+9.8•5.4= 40.54 m/s
At the top point v=0
0 =v₀-gt
t = v₀/g=40.54/9.8 =4.14 s.
t₃=t₂- t=5.4 – 4.14 = 1.26 s
v₂=gt₃ = 9.8•1.26 = 12.38 m/s (directed dowmward)