Enter the initial height, launch angle, and velocity of the projectile to calculate its time of flight.
Use this calculator to determine the time of flight for a projectile as long as it remains in the air after launch. Get complete steps in the solution as well.
Here arise a couple of cases, which are as follows:
If Projectile Thrown From Ground (Given Initial Height = 0)
By using 2nd equation of motion,
$$ y = V_{o}t sin\left(\alpha\right) - \dfrac{1}{2}*g*t^{2} $$
When the object hits the ground, the vertical component (y) becomes zero, such that;
$$ 0 = V_{o}t sin\left(\alpha\right) - \dfrac{1}{2}*g*t^{2} $$
$$ -V_{o}t sin\left(\alpha\right) = - \dfrac{1}{2}*g*t^{2} $$
$$ -V_{o} sin\left(\alpha\right) = - g*\dfrac{t^{2}}{t} $$
$$ t = \dfrac{V_{o} sin\left(\alpha\right)}{g} $$
If Projectile Thrown From Certain Height (Given Initial Height > 1)
$$ t = \dfrac{V_{o}sin\left(\alpha\right) + \sqrt{\left(V_{o}sin\left(\alpha\right)\right)^{2} + 2gh}}{g} $$
Where;
Find the time of flight for a projectile with the following values:
$$ α = 75^{o} $$
$$ \text{Initial Height} = 20 meters $$
$$ \text{Initial Velocity} = 50 km/h $$
$$ Gravitational Force = g = 9.8 m/s^{2} $$
Using the time of flight formula:
$$ t = \dfrac{V_{o}sin\left(\alpha\right) + \sqrt{\left(V_{o}sin\left(\alpha\right)\right)^{2} + 2gh}}{g} $$
$$ t = \dfrac{ 50 \times sin(75) + \sqrt{(50 \times sin(75))^2 + 2 \times9.80665\times20}}9.80665 $$
$$ t = \dfrac{ 50 \times 0.9659 + \sqrt{(50 \times 0.9659)^2 + 2 \times9.80665\times20}}9.80665 $$
$$ t = \dfrac{ 13.42 + \sqrt{(13.42)^2 + }}9.80665 $$
$$ \text{t = 3.807 sec} $$