Derivation of the Maximum Height Reached by a Projectile Tbottom = mgH1 + 1/2 mV02 Energy is Conserved Ttop = mgH2 + 1/2 m(Vtop)2 mgH1 + 1/2 mV02 = mgH2 + 1/2 m(Vtop)2 H1 = 0 1/2 mV02 = mgH + 1/2 m(Vcosq )2 Divide by m; m cancels 1/2 V02 = gH + 1/2 (Vcosq )2 V2 = 2gH + (Vcosq )2 Multiply both sides by 2 1/(cos2q ) = (2gH)/ (Vcosq )2 + 1 Divide both sides by V2 = 2gH + (Vcosq )2 (Vcosq )2 (V2(1 - cos2q) / (2g) = H Multiply both sides by (Vcosq )2 Energy is Conserved Ttop = mgH2 + 1/2 m(Vtop)2 mgH1 + 1/2 mV02 = mgH2 + 1/2 m(Vtop)2 H1 = 0 1/2 mV02 = mgH + 1/2 m(Vcosq )2 Divide by m; m cancels 1/2 V02 = gH + 1/2 (Vcosq )2 V2 = 2gH + (Vcosq )2 Multiply both sides by 2 1/(cos2q ) = (2gH)/ (Vcosq )2 + 1 Divide both sides by V2 = 2gH + (Vcosq )2 (Vcosq )2 (V2(1 - cos2q) / (2g) = H Multiply both sides by (Vcosq )2 H = (V2(1 - cos2q) / (2g)