A farm truck travels due east with a constant speed of 9.50 m/s along a horizontal road. A boy riding in the back of the truck tosses a can of soda upward (Fig. P3.40) and catches it at the same loca- tion in the truck bed, but 16.0 m farther down the road. Ignore any effects of air resis- tance. (a) At what angle to the vertical does the boy throw the can, relative to the moving truck? (b) What is the can’s ini- tial speed relative to the truck? (c) What is the shape of the can’s trajectory as seen by the boy? (d) What is the shape of the can’s trajectory as seen by a stationary observer on the ground? (e) What is the initial velocity of the can, relative to the stationary observer?

a)he angle is 90 with respect to the horizontal (x axis)

b) its speed is zero both vertically and horizontally

c)  vertical path

d)  a parabolic movement

e)  v₀ = (9.5i ^ + 8.232 j ^) m / s

Explanation:

This is a relative problem and movement.

            .v ’= v + u

Where v ’is the speed with respect to the mobile system, v the speed with respect to the fixed system and the speed between the two reference systems

a) The child and the can is in the truck, so they go at the speed of the truck, when he throws the can he continues at this speed on the x-axis and therefore as the two advance the same distance the more hands of the child, consequently the can is thrown vertically

The angle is 90 with respect to the horizontal (x axis)

b) with respect to the truck, the can is still, so its speed is zero both vertically and horizontally

c) The child sees that the can follows a vertical path

d) A stationary observer on the ground, sees that the can has a constant speed in the same direction of the truck and when they throw it vertical goal has a vertical movement, the sum of these two movements gives a parabolic movement of the same uncle as a projectile launch

e) the initial speed has two components

X Axis         v_lata = v_camion = 9.5 m / s

Y Axis          speed given by the child

Let's look for the travel time

         x = v₀ₓ t

         t = x / v₀ₓ

         t = 16 / 9.5

         t = 1.68 s

         y = t - ½ g t²

When he returns to the child's hand and = 0

          0 = v_{oy} t - ½ g t²

          v_{oy} = ½ g t

          v_{oy} = ½  9.8  1.68

          v_{oy} = 8,232 m / s

Speed is

         v₀ = (9.5i ^ + 8.232 j ^) m / s

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

c) up

hope this