As a car travels down the highway at 25.0 m/s, a passenger tosses a can out the

Question

As a car travels down the highway at 25.0 m/s, a passenger tosses a can out the window at an angle of 45 degree above the horizontal and perpendicular to the motion of the car. The initial speed of the can relative to the car is 1.0 m/s. The can is released 1.20 m above the road surface. Choose the origin of your coordinate system to be the point on the road surface directly below the release point of the can. Choose your positive x-axis to be in the direction of the motion of the car. Choose your positive y-axis to lie in the horizontal plane and directed to the left of the positive x-axis. Choose the positive z-axis to be up. What is the x component of the can's release velocity? What is the y component of the can's release velocity? What is the z component of the can's release velocity? How long is the can in the air? Where does the can land, relative to the release point along the road?

Explanation / Answer

(a)
X component of the Can's release velocity = 0

(b)
Y component of the Can's release velocity = 0

(c)
Z component of the Can's release velocity = 1.2 m

(d)
Velocity in z direction, = 10 * sin(45) = 7.07 m/s

Distance s = 1.2 m

s = u*t + 1/2*a*t^2
-1.2 = 7.07 * t - 1/2*9.8*t^2
t = 1.6 s

Time can is in the air, t = 1.6 s

(e)
Speed of can relative to the road, = 25 + 10 * cos(45)
v = 32.07 m/s

Distance travelled in 1.6 s, = 32.07 * 1.6 m
Distance travelled  = 51.3 m

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