Is my formulae correct for time of flight in this projectile problem?

[Benjamin_harsh]

Homework Statement

Is my formulae correct for time of flight in this projectile problem?

Relevant Equations

Is my formulae correct for time of flight in this projectile problem?

$V_{oy} = V_{0}.sinθ = (20 m/sec)(sin 30) = 10 m/sec$

$V_{ox} = V_{0}.cosθ = (20 m/sec)(cos 30) = 17.3 m/sec$

$y = y_{0} + V_{oy}t - \large\frac {1}{2}\normalsize gt^{2}$

$0 = 50 + 10t - 4.9t^{2}$

$0 = 4.9t/6{2} - 10t - 50$

$t = \large\frac {-b±\sqrt {b^{2} - 4ac}}{2a}$

$t = \large\frac {10 ±\sqrt {100 - 4(4.9)(-50)}}{9.8}$

$t = \large\frac {10 ± 32.86}{9.8}\normalsize = 4.37 sec \;(or) -2.33 sec$

$Y_{max}$; Max height = $\large\frac {V^{2}sin^{2}θ}{2g} + \normalsize H = \large\frac {100}{19.6} \normalsize + H = 55.102m$

My formulae = I simple add these two time intervals 1) Time it takes from A to travel to B;

$T_{A → B} = \large\frac{V_{0}.sinθ}{g}$

$= \large\frac{10}{9.8} \normalsize = 1.0204 sec.$

2) Time it takes for for B to travel to C; $T_{B→C} =\large\sqrt\frac {2.y_{max}}{g}$

$= \large\sqrt\frac {2.(55.102)}{9.8}\normalsize = 3.353 sec$$T_{A → B} + T_{B→C} = 4.37 sec$

Is my formulae correct for time of flight in this projectile problem?

 

[scottdave]

It looks ok to me. Just there's a typo in your latex, right before the quadratic formula. It should be 4.9 t squared rather than 4.9 t/62

 

[Ssnow]

Yes, you must solve the second degree equation and discard the negative solution ... it seems correct and you proved it calculating $T_{A\rightarrow B}$, $T_{B\rightarrow C}$ and verifying that the sum is equal to $4.37 s$.Ssnow