Physics EK 1001 #118 - Shorter method?

[MAnne]

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Would something like this be on the MCAT? They say to find airtime first then multiply it by the horizontal velocity. The answer to this question seems like it would take so long. I understand how they got the answer but I didn't know how to approach it when I first looked at it. Can anyone help please?

A projectile is launched at an angle of 30º to the horizontal from a 15m platform. Its initial velocity is 20 m/s. How far does it travel?

Thanks.


~M

 

[milski]

No real shortcuts here, sorry. Get the horizontal and vertical components of its velocity, calculate how much time it will take for it to fall to the ground, calculate horizontal distance from horizontal velocity and time.

It's really not that horrible once you start working through it. Sometimes doing the work can turn out to be faster than trying to come up with a shortcut.


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[lastdrop]

While I understand how they approach the problem in their explanation, I am struggling to understand one part.

The initial height is 15m, and we need to know the maximum height. To calculate this, EK explains they use x = 1/2at^2, but wouldn't using that equation only be correct if the initial velocity is 0?

The question states that the initial velocity is 20m/s @ 30 degrees, so the vertical velocity would be 20sin(30) = 20*0.5 = 10m/s.

They conclude that the maximum height reached is 20 (5 from the equation x=1/2at^2 plus 15 due to the platform).

Am I missing something?

 

[Sports Junky]

While I understand how they approach the problem in their explanation, I am struggling to understand one part.

The initial height is 15m, and we need to know the maximum height. To calculate this, EK explains they use x = 1/2at^2, but wouldn't using that equation only be correct if the initial velocity is 0?

The question states that the initial velocity is 20m/s @ 30 degrees, so the horizontal velocity would be 20sin(30) = 20*0.5 = 10m/s.

They conclude that the maximum height reached is 20 (5 from the equation x=1/2at^2 plus 15 due to the platform).

Am I missing something?

First, the horizontal velocity is not 10 m/s, the vertical velocity is 10 m/s. Using Vf = Vi + at, you can calculate the time to reach the top of the projectory to be 1 second. The average velocity during this period is 5 m/s, so the height it traveled must have been 5m.
At this point, you could use the free-fall equation to calculate the remaining time (2 seconds).

Putting this altogether, the horizontal distance traveled is simply 20*cos(30)*3s ~52m.