A car is travelling north on a highway at 25 m/s. Just as the car crosses a perpendicularly intersecting crossroad, the passenger throws out a can horizontally, towards the east. The initial speed of the can relative to the car is 10 m/s. It is released at a height of 1.2 m above the road.
a) What is the initial velocity of the can relative to the road?
b) Where does the can land?
I mainly having trouble figuring the first question. If you could explain it that would be great!
the can is moving 10 m/s from the car
... while the car is moving 25 m/s on the road
add the two vectors to find the can's velocity relative to the road
the can lands north and east of the crossroads
... use the release height to find the flight time
a. Vo = 10 + 25i. = 26.93m/s[68.2o] = Initial velocity.
b. h = 0.5g*t^2 = 1.2,
4.9*t^2 = 1.2,
t = 0.495 s. = Fall time.
d = Vo*t = 26.93[68.2o] * 0.495 = ___m[68.2o]
I got the exact same question (see #6) but with 25 m/s changed to 12.5 m/s and 1.2 m changed to 1.6 m.
You can see the work I showed and swap out the appropriate numbers.
aquantaday.wordpress.com/grade-12-physics-challenge-problems-1-kinematics/
To find the initial velocity of the can relative to the road, we need to consider the velocities of both the car and the can and combine them appropriately.
The car is traveling north at 25 m/s, which means its velocity can be represented as 25 m/s upward on a vertical coordinate system. Meanwhile, the can is thrown horizontally towards the east with an initial speed of 10 m/s.
In order to combine these velocities, we can use vector addition. On a coordinate system, we can define the upward direction as the positive y-axis and the eastward direction as the positive x-axis.
The initial velocity of the can relative to the road can be represented as a vector sum of the car's velocity (25 m/s upward) and the can's velocity (10 m/s to the right).
The magnitude of the initial velocity of the can relative to the road can be found using the Pythagorean theorem. We can calculate it as follows:
velocity of the can relative to the road = sqrt((velocity of car)^2 + (velocity of can)^2)
velocity of the can relative to the road = sqrt((25)^2 + (10)^2) m/s
velocity of the can relative to the road = sqrt(625 + 100) m/s
velocity of the can relative to the road = sqrt(725) m/s
Therefore, the initial velocity of the can relative to the road is approximately 26.94 m/s.