Answer to Question #94860 in Mechanics | Relativity for stefanie

Question

A farm truck moves due east with a constant velocity of 10.00 m/s on a limitless horizontal stretch of road. A boy riding on the back of the truck throws a can of soda upward (see figure) and catches the projectile at the same location on the truck bed, but 20.0 m farther down the road.
(a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can?(b) What is the initial speed of the can relative to the truck?(in m/s)(e) In this observer's frame of reference, determine the initial velocity of the can.(magnitude and direction in degree above horizontal)

Answer 1

Let's choose the x-axis horizontally in the direction of movement of the machine and y-axis vertically upward with the (0,0) point at the initial position of the projectile and rigidly attach it to this point on farm truck (reference system of the truck). Let's denote $\vec v = ({v_x},{v_y})$ – the initial speed of the projectile, $\vec u = ({u_x},0) = (10,0)$ – speed of the truck, $l$ – the distance travelled by the truck while projectile was in the air. Also in this problem we shall use $g = 10[\frac{m}{{{s^2}}}]$.

a)

"The boy catches the projectile at the same location" means that in the frame of reference of the truck the projectile didn't move along x-axis (in the horizontal direction) thus it was thrown vertically upwards (along y-axis) and the required angle to the verical (y-axis) is equal to zero, $\alpha = 0[{\text{rad}}]=0[^\circ ]$

Answer: the angle between initial speed and the vertical in frame of reference of the truck in equal $0[^\circ ]$

b)

First, find the time that the projectile was in the air using uniform movement of the truck

$$t = \frac{l}{{{u_x}}} = \frac{{20[m]}}{{10[\frac{m}{s}]}} = 2[s]$$

Using the result of the a) we get $\vec v = (0,{v_y})$ - no movement along x-axis and using law for uniformy accelerated motion (in Earth gravitational field)

$${v_{y}}t - \frac{g}{2}{t^2} = 0$$

where ${v_{y}}$ is the projection of the initial velocity of the projectile on the y-axis. Expressing ${v_{y}}$ we get

$${v_{y}} = \frac{1}{2}gt = \frac{1}{2} \cdot 10[\frac{m}{{{s^2}}}] \cdot 2[s] = 10[\frac{m}{s}]$$

Answer: initial velocity of the projectile relative to the truck is equal to ${{\vec v}} = (0,10)[\frac{m}{s}]$ (the initial speed is ${v_{y}} = 10[\frac{m}{s}]$)

c)

To go to the observer's frame of reference we can use Galilean transformation (we shall use prime-mark to for quantities in the observer's frame of reference) and we get the initial velocity in this frame

$$\vec v' = \vec v + \vec u = (0,10)[\frac{m}{s}] + (10,0)[\frac{m}{s}] = 10(1,1)[\frac{m}{s}]$$

To get the magnitude we should calculate the norm of this vector

$$\left| {\vec v'} \right| = 10\sqrt {{1^2} + {1^2}} [\frac{m}{s}] = 10\sqrt 2 [\frac{m}{s}]$$

To get the direction (angle between initial velocity and x-axis in observer's frame of reference) we can use scalar product between ${\vec v'}$ and unit vector along x-axis $\vec i = (1,0)$

$$\cos \theta = \frac{{\vec v' \cdot \vec i}}{{\left| {\vec v'} \right|\left| {\vec i} \right|}} \approx \frac{{10[\frac{m}{s}] \cdot 1 + 10[\frac{m}{s}] \cdot 0}}{{10\sqrt 2 [\frac{m}{s}] \cdot 1}} = \frac{1}{{\sqrt 2 }}$$

And finally

$$\theta = \arccos \frac{1}{{\sqrt 2 }} = \frac{\pi }{4}[{\text{rad}}] = 45[^\circ]$$

Answer: the magnitude of the initial velocity of the projectile in the observer's frame of reference is equal to $\left| {\vec v'} \right| = 10\sqrt 2 [\frac{m}{s}]$, the direction is equal to $\theta = 45{[^\circ }]$