A. Vˉ=f(4)−f(0)4=30.4mc\bar{V}= \frac{f(4)-f(0)}{4}=30.4 \frac{m}{c}Vˉ=4f(4)−f(0)=30.4cm
B. V(t)=ft′=−9.8t+50V(t)=f_t'=-9.8t+50V(t)=ft′=−9.8t+50 hence V(4)=10.8msV(4)=10.8\frac{m}{s}V(4)=10.8sm
C. a(t)=Vt′=−9.8ms2a(t)=V_t'=-9.8\frac{m}{s^2}a(t)=Vt′=−9.8s2m
D. Ifτ\tauτ is time up then V(τ)=0V(\tau)=0V(τ)=0 hence −9.8τ+50=0-9.8\tau+50=0−9.8τ+50=0 then τ=5.1s\tau=5.1sτ=5.1s then f(τ)=H≈132.6mf(\tau)=H\approx132.6mf(τ)=H≈132.6m
E. If ttt is time down then H=gt22H=\frac{gt^2}{2}H=2gt2 hence t=2Hg≈5.1st=\sqrt \frac{2H}{g}\approx5.1st=g2H≈5.1s then T=t+τ≈10.2sT=t+\tau\approx10.2sT=t+τ≈10.2s