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 |
H = (V2(1 - cos2
q) / (2g) HOME