Projectile Motion & Motorcycle Jumps

AI Thread Summary
The discussion revolves around understanding projectile motion in the context of a motorcycle jump across a canyon. The user seeks clarity on calculating the minimum speed and time of flight for a jump, given a distance of 15 meters and an incline angle of 37 degrees. Confusion arises from the equations used, particularly the significance of setting the height equation to zero, which indicates the projectile's return to the same vertical position. The conversation highlights the use of different gravitational constants, with some participants suggesting a simplified value of 10 m/s² instead of the more precise 9.8 m/s². The forum provides insights into solving the problem by equating the initial and final heights to zero to determine the time in the air.
amyalex
Messages
2
Reaction score
0
Hi:
This is my first visit to this website. I am a 43-year-old undergraduate majoring in Environmental Horticulture and somehow missed physics along the way. It's been 2-3 years since I had all my algebra, trig and pre-calc and now in my first 2 weeks of physics I find I need help. We just studied projectile motion and the problems that go with it. I think I understand the concepts and can find the answer to a problem that deals with how far the projectile travels given an inital velocity and angle. However, the problem I'm trying to do now the distance is given and I need to come up with the time and the velocity. I actually have my instructor's notes how to solve this, but his notes seem more convoluted than what it necessary based on other websites I have seen. Also, I notice that everyone uses different formulas or at least different symbols in those formulas and it is all very confusing. At any rate, the problem is A daredevil jumps a canyon 15 m wide by driving a motorcycle up an incline sloped at an angle of 37 degrees with the horizontal. What minimum speed must she have in order to clear the canyon? How long will she be in the air? He starts out with the horizontal being Vo = 15/.8t and I'm ok with that until later. Then for the vertical, working from the equation y = Voyt - 1/2 a(or g in this case)t^2, my professor works to Vo sin 37 deg. t - 5t^2 = 0. Here's where I get confused. Does the 5 in the 5t^2 come from 1/2 of 10 m/s for g and how does this equation equal 0 and what is the significance of that? He then goes on to substitutions and other things that I pretty much understand, but I can't get past this zero business.

Thanks for your help and this probably won't be my last post this semester!

Amy
Gainesville, Florida
 
Physics news on Phys.org
Yes 5=10/2

Okay what is the displacement in y direction
 
A more accurate figure for the "acceleration due to gravity" is 9.8 m/s2 but 10 is okay for an example. I wouldn't want to be that "rough" in planning how to jump a canyon but I guess making the initial speeder higher than necessary is the way to go!
 
Thanks for answering my question, but each of you has just answered part of it. Can you answer the other part of my question, because that's the part I'm really having trouble with. And by the way, the professor is having us use 10 m/s^2 instead of 9.8.
 
That equation gives the height of the projectile at time t.

The height is 0 at the beginning and at the end of the flight. To find out how long the motocycle was in the air, you can calculate the time it takes for the equation to be zero.
 
The basic formula is
y-y_0=v_ot+\frac{at^2}2

here y=position of particle after time t
y0= initial position
here a=-g

Now after landing the other side of the cannon the position will be same as initial therefore y=y0

Hence u get y-y0= 0
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »
Thread 'Minimum mass of a block'

Thread 'Minimum mass of a block'

Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
View full post »

Similar threads

Find the height of a lamp based on the velocities of projectiles
Replies
40
Views
2K
I need opinions on a Projectile Motion problem that I made up
Replies
11
Views
2K
Find Projectile Flight Time Given Only Maximum Height
Replies
4
Views
2K
How Do You Solve a Projectile Motion Problem?
Replies
3
Views
4K
Attempt to jump across a river on a motorcycle
Replies
2
Views
3K
Relationship between horizontal range and angle of launch
Replies
7
Views
3K
Why does this not work? Projectile Motion problem for Dynamics Class
Replies
19
Views
3K
Projectile Motion: time it takes for projectile to hit truck
Replies
15
Views
3K
Projectile Motion Experimental Error
Replies
1
Views
2K
Calculating Forces on a Motorcycle Riding in a Transparent Sphere
Replies
5
Views
2K

Hot Threads

Recent Insights

Back
Top