In the previous example of time of flight, the out of control bus is forced to jump from a location $\langle 0,40,-5 \rangle$m with an initial velocity of $\langle 80,7,-5 \rangle m/s^-1$. We have now found the time of flight to be 9.59s and now want to find the position of where the bus returns to the ground.

• Starting position of the bus $\langle 0,40,-5 \rangle$
• Initial velocity of the bus $\langle 80,7,-5 \rangle$
• The acceleration due to gravity is 9.8 $\dfrac{m}{s^2}$ and is directed downward.
• The bus experiences one force - the gravitational force (directly down).
• The bus takes 9.59s to reach the ground (from previous problem link)

• The final position of the bus.

• Assume no drag effects
• Assume ground is when position of bus is 0 in the y direction

Diagram of situation.confused by what's labeled

Diagram of forces acting on bus once it leaves the road.

Equation for calculating the final position of an object.

$$ x_f = x_i + V_{avg,x} \Delta{t}$$

From the previous problem you already know the final location of the ball in the y direction to be 0 as it has met the ground after 9.59s.

Now to find the range in the x and z directions:

$$ x_f = x_i + V_{avg,x} \Delta{t}$$

$$ = 0 + 80m/s(9.59s)$$

$$ = 767m$$

$$ z_f = z_i + V_{avg,z} \Delta{t}$$

$$ = -5 + -5m/s(9.59s)$$

$$ = -52.95$$

Final position = $\langle 767,0,-52.95 \rangle$ m