# An object is thrown vertically from a height of 9 m at 43 m/s. How long will it take for the object to hit the ground?

Jan 6, 2017

$\Delta {t}_{t o t} \approx 9 s$

#### Explanation:

The fall time and rise time will be different as the object begins and ends its motion at different altitudes. Ignoring air resistance, you can use kinematic equations to calculate the total flight time.

For the rise time:

${v}_{f} = {v}_{i} + a \Delta {t}_{r i s e}$

The object is launched at an initial velocity (${v}_{i}$) of $43 \frac{m}{s}$. When it reaches its maximum altitude, it will pause momentarily, changing direction, before falling back to the ground. Therefore, for this equation, ${v}_{f} = 0$. The acceleration for an object in free-fall is equal to $- g$, the free fall acceleration, $- 9.8 \frac{m}{s} ^ 2$.

Solve for $\Delta {t}_{r i s e}$:

$\Delta {t}_{r i s e} = \frac{\cancel{{v}_{f}} - {v}_{i}}{a}$

$\Delta {t}_{r i s e} = \frac{- 43 \frac{m}{s}}{- 9.8 \frac{m}{s} ^ 2} = 4.4 s$

If the object began and ended at the same altitude, we could simply multiply this result by two to get the total flight time. However, the object begins at $9 m$ and ends at $0 m$, on the ground. We can use a kinematic to calculate the maximum altitude of the projectile to find how far it travels upwards from its initial position of $9 m$. Adding $9 m$ to this altitude, we will get the total distance that the object falls from the top of its trajectory.

${\left({v}_{f}\right)}^{2} = {\left({v}_{i}\right)}^{2} + 2 a \Delta y$

$\Delta y = \frac{- {\left({v}_{i}\right)}^{2}}{2 a}$

$\Delta y = \frac{- {\left(43 \frac{m}{s}\right)}^{2}}{2 \cdot - 9.8 \frac{m}{s} ^ 2}$

$\Delta y = 94.34 m$

This is the maximum altitude of the object. Adding $9 m$ to this, we have that the object falls a total vertical distance of $103.34 m$. That is, it travels $- 103.34 m$ vertically. Now we can use a final kinematic to calculate the fall time. Note that the initial velocity of the object is now zero, as it begins from its maximum altitude this time.

$\Delta y = \cancel{{v}_{i} \Delta {t}_{f a l l}} + \frac{1}{2} a {\left(\Delta {t}_{f a l l}\right)}^{2}$

Solve for $\Delta {t}_{f a l l}$

$\Delta {t}_{f a l l} = \sqrt{\frac{2 \Delta y}{a}}$

$\Delta {t}_{f a l l} = \sqrt{\frac{2 \left(- \left(94.34 + 9\right)\right) m}{- 9.8 \frac{m}{s} ^ 2}}$

$\Delta {t}_{f a l l} = 4.59 s$

We can add these times together to get the total flight time:

${t}_{t o t} = {t}_{f a l l} + {t}_{r i s e} = 4.59 s + 4.4 s = 8.99 s$

This is $\approx 9 s$.