# You are in the back of a pickup truck on a warm summer day and you have just finished eating an apple. The core is in your hand and you notice the truck is just passing an open dumpster 7.0 m due west of you. The truck is going 30.0 km/h due north - cont?

## and you can throw that core at 60.0 km/h. In what direction should you throw it to put it in the dumpster, and how long will it take it to reach its destination?

Feb 26, 2016

My vantage point in the truck:
$v \left(t\right) \approx 60 j - 10 \cdot \frac{7}{10} k = 60 j - 7 k$ I am rounding $g \to 10$
$t i m e , t = \frac{7}{10} s$

$v \left(t\right) = {v}_{x} i + {v}_{y} j - \text{gt} k$
v_(x)hatx+v_yhaty - "gt"hatz = ((v_x),(v_y),("-gt")) = ((-30),(60),("-9.81t")) or

4) $v \left(t\right) = - 30 i + 60 j - 7 k$
The direction is given in the x-y plane is give by the angle between
the vector given by (-30i + 60j); theta = tan^-1(-2) = -63.4^0 or ${296.5}^{0}$

Remark: You can also use conservation of momentum to get the direction. I have added the z direction because the core will influenced by gravity, thus will undergo a parabolic motion as it travels to dumpster...

Observer outside the truck vantage point

#### Explanation:

This is a great question that illustrate relative displacement and velocity, or in general acceleration. While your question does not touch on it the general consideration of this is to determine the ball
trajectory in the presence ${v}_{y} , - {v}_{x} \text{ and } {a}_{z} = g$. I will try to give you insight into both the simplified 2-D and 3-D views of the problem. I will do this from my reference point in the truck (which is what your question asks) and from an observer outside the train.

Observer - Inside the truck, Me : The core will move with the constant velocity, ${v}_{\text{North}} = {v}_{y} = 60 \frac{m}{s}$ away from the train. There is nothing that slows the core. So I will see the ball right in front of me, flying further away and falling down with v_z=−gt
obviously, there will be a curved trajectory, a parabola in the y-z, the plane where the train is moving perpendicularly to. So what I see is the vector,
1) v(t) =v_yj - "gt"k = v_yhaty - "gt"hatz = ((0),(v_y),("-gt")) = ((0),(v_y),("-9.81t")) or
2) $v \left(t\right) = 60 j - 9.81 t k$
To calculate t, you use the ${v}_{y}$ and the distance to the dumpster
distance $y = 7 m$
$t = \frac{7 m}{60 \frac{m}{s}} = \frac{7}{60} s \approx .1167$ insert this in 2 and we have:
3) $v \left(t\right) \approx 60 j - 10 \cdot \frac{7}{10} k = 60 j - 7 k$ I am rounding $g \to 10$

Observer - Outside the truck, You clearly an observers on the side walk close to the truck will also see the speed of the truck so we need to adjust equation 1) and 2) as:
3) $v \left(t\right) = {v}_{x} i + {v}_{y} j - \text{gt} k$
v_(x)hatx+v_yhaty - "gt"hatz = ((v_x),(v_y),("-gt")) = ((-30),(60),("-9.81t")) or
4) $v \left(t\right) = - 30 i + 60 j - 7 k$

The direction is given in the x-y plane is give by the angle between
the vector given by (-30i + 60j); theta = tan^-1(-2) = -63.4^0 or ${296.5}^{0}$