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

• Benjamin_harsh
In summary: Snowy: In summary, using the given information, we can calculate the time it takes for the projectile to travel from point A to B and from point B to C. By adding these two time intervals, we can determine the total time of flight for the projectile. Additionally, we can use the given formula to calculate the maximum height reached by the projectile. Lastly, the formula used to calculate time of flight appears to be correct as it was used to verify the calculated time intervals.
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

Benjamin_harsh
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

Benjamin_harsh

1. What is the formula for time of flight in a projectile problem?

The formula for time of flight in a projectile problem is t = 2v0sin(θ)/g, where t is the time of flight, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

2. How do I know if my formula for time of flight is correct?

To determine if your formula for time of flight is correct, you can compare it to the standard formula mentioned above. If they are the same, then your formula is correct.

3. Can I use the same formula for time of flight in all projectile problems?

Yes, the formula for time of flight is applicable to all projectile problems, as long as the initial velocity and launch angle are known.

4. What units should I use for the variables in the time of flight formula?

The units for time of flight will depend on the units used for initial velocity and acceleration due to gravity. Typically, initial velocity is measured in meters per second (m/s) and acceleration due to gravity is measured in meters per second squared (m/s2). Therefore, the time of flight will have units of seconds (s).

5. What if my projectile problem involves air resistance? Can I still use the same formula for time of flight?

In cases where air resistance is a factor, the formula for time of flight will need to be modified to take into account the effects of air resistance. This will require additional information, such as the mass and drag coefficient of the projectile, and the air resistance formula will need to be incorporated into the time of flight formula.

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