Question
Imagine that you are in a car that is traveling up a bridge that is inclined at an angle of 19.0 degrees from the horizontal. The car's speed is 56.0 mi/hr (25.03 m/s). At time t = 0 you open your window and illegally drop a soda can from your rest frame over the edge of the bridge. Your height above ground level is 43.0 m. How far does the can travel horizontally before hitting the ground? Neglect air resistance.
At what angle does the can strike the ground (in degrees)? This is the angle which the can's velocity vector makes with the horizontal just before hitting the ground.
What is the speed of the can just before it hits the ground?
Explanation / Answer
First thing we need to assume is that, when the driver lets go of the soda, the soda has acquired an intial velocity that is similar to that of the car. So by resolving the horizontal and vertical components of the velocity; 25.03cos19 and 25.03sin19 respectively. Vertically, the displacement of the soda would be 43.0m. Thus, s = ut +0.5at^2 -43 = (25.03sin19)t - 0.5gt^2 (Taking upwards as positive) 0.5gt^2 - (25.03sin19)t - 43 = 0 This is quadratic equation: t = 3.91 sec or t = -2.25 sec (rejected since time has not negativity.) Now that you know the total time taken for the soda to travel, you can solve the horizontal distance it has travelled: s = ut s = (25.03cos19) x 3.91 s = 92.5m For the next part, you need to find the vertical component of the velocity just before the soda hits the ground. Again, since you know the total time taken, applying law of kinematics, v = u + at v = (-25.03sin19) + (9.80)(3.91) (Taking downwards as positive) v = 30.17m/s Horizontal component of velocity remains unchanged (no acceleration acting on it) = 25.03cos19 = 23.67m/s Thus, to find the angle: tan? = 30.17/23.67 ? = 51.9 degrees For the last part, the speed the question is asking for is the summation of the horizontal and vertical components of velocity. Thus, speed = ((30.17)^2 + (23.67)^2)^0.5 speed = 38.3m/s