Physics 3.5.4e - Projectile Practice Problem 5

Practice Problem on Projectile Motion.

[skylar]

I wish the questions I'm being asked were as straight forward as these are. They're complicated enough to make you think, but not easy enough to figure out without understanding the concepts. I'm so screwed in physics. The online homework is like harvard level questions, versus what the textbook teaches, and the professor explains. I wish I would have been set on the right path from the start. I'm at least 80 hours into trying to figure out all of this, and am still totally unsure of how to approach almost every problem I come across. Every single question has a curveball. Your videos are a light guiding me in the right direction tho, it's been dark af for a long time. Thank you for creating so many practice problems, with great explanations.

[TheMiAStudios]

In this instance, if you find peak height you get something around 12m which isn't high enough to go over the wall. And that is the peak height meaning it never goes over. From that, you can infer the same that you got, no it does not clear the wall.

[derekowens]

@WinsonLaw93 Good question. We can't use the peak height to determine if it will clear the wall because it is not at its peak height vertically at the same time it is horizontally at the wall. What he have to find is whether it is above the height of the wall at the moment that it is at the horizontal position of the wall. And that is a separate question from What is the peak height.

[derekowens]

@WinsonLaw93 Remember that you have to treat the horizontal motion independently from the vertical motion. So we first find the time it takes to go *horizontally* to the wall. Then we find whether or not it is, at that moment in time, above the wall, vertically. Hope that helps, D.O.

[derekowens]

@albasser89 I think your formula will give you the total time, and if you use this time to find the horizontal distance, then you find that it does reach the wall. The question, though, is whether or not it is high enough to clear the top of the wall at the time that it reaches the wall, so you need to approach it differently.

[darpanpathak]

So I used v_f = v_0 + at and I got t = 1.586 s, then I plugged it into

y_f = y_0 + v_0*t + ((1/2)at^2) and I got 12.34 m, which means the rock still did

not go over the wall, but how to know which equation to use. x_f=vt or v_f = v_0 +at

[holinatoliman]

Please can someone do this question as an example...(a car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal.The negligent driver leaves the car in neutral, and the emergency brakes are defective.The car rolls from rest down the incline with a constant acceleration of 4 meter per squared seconds for a distance of 50m to the edge of the cliff, which is 30m above the ocean. find the length of time the car is in air.(b) find the car's position relative to the base of the cliff when the car lands in the ocean..

[reigncherrishreyes]

I used the height to get the time, and my range is over 45m, so does it mean that the rock can pass the wall depending on the solution?

[pasado]

I have a question. Why did you use 2.89 s directly to solve for the height of the rock at that time? Wouldn't that be not correct? The rock will first go upward before going downward, so it will use different signs of gravitational acceleration. I'm confused.

[Bria.dances]

Thanks for the practice, but why didn't the rock reach over the building? Is it because the distance if the rock is 4.1m ?

[Ahmad-rbtm]

For anyone who has seen the video recently.. For the distance, while calculating the time, he used 45 meters. That's the distance from the catapult to the wall, is it right just assume that's also the distance the rock travels (what he did in the video) although we haven't been given that? I started with the vertical component and used the Vf=Vi+at and got the time=1.6 seconds instead of what he got (2.89). Am I making a mistake here ?

[TehhStarzHD]

u didnt start from a height above the ground. lets say i started 5 m above the ground. Would i add 5 to teh 4.04 ?

[shrujalagrawal]

you are amazing!!!! Please make videos on the chapter uncertainties if possible

[miodf]

then how would you get the total time traveled if its not 2.89s?

[Gsmstfreshie]

5:56. you lost me there.. doesn't the the total time of the flight equals to the total time it took for the distance to go horizontally? @.@

help!

[derekowens]

@Denseworldproduction Maybe. Tell me what info you are given, or just tell me the whole problem.

[kyubey]

You saved me in calculus based physics

[ColenzJoachimCamero]

is my method correct? since the wall is 20 meters high, i just solve for the maximum height of the projectile which is 12.33 meters and just by seeing the maximum height I can assume that it will not pass through the wall.

[persianposer]

I understand the equation for the Yf would be :
Yf=Yo+Vyit - 1/2gt^2   you added 1/2 gt^2 instead of subtracting it. Why

[Atizzleify]

When I used trig to find the x and y components, I definitely did not get a full decimal number. I got 11sqrt(2)