Example: Finding the range of a projectile

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 3.65s and now want to find the position of where the bus returns to the ground.

Facts

Lacking

Approximations & Assumptions

Representations

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

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The general equation for calculating the final position of an object:

$$ {\overset{\rightarrow}{r}}_{f} = {\overset{\rightarrow}{r}}_{i} + {\overset{\rightarrow}{v}}_{avg}\Delta t $$

Also know as the position update formula.

Solution

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.

We now to find the range in the x and z directions in order to have a position vector for the final resting place of the bus.

There is no force acting in the x or z directions as the only force acting on the system is the gravitational force which acts in the y-direction.

This means that the initial velocities in both of these directions have remained unchanged.

We know the amount of time the bus has been traveling in the x-direction at its initial velocity and its initial position so we can compute the distance travelled in this direction using the position update formula for x-components.

$$ x_{f} = x_{i} + V_{avg,x}\Delta t $$

Plug in respective values for variables.

$$ = 0 + 80m/s(3.65s) $$

Compute range in x-direction.

$$ = 292m $$

Repeat same process for the z-components:

$$ z_{f} = z_{i} + V_{avg,z}\Delta t $$

Plug in respective values for variables.

$$ = - 5 + - 5m/s(3.65s) $$

Compute range in z-direction.

$$ = - 23.25m $$

Write range(final position vector) using all components:

Final position =

$$ \langle 292,0, - 23.255\rangle m $$