fastnfuriouseng
Badges: 5
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 11 months ago
#1
Hiya! I have been attempting a level 5 kinematics problem but don't seem to be getting anywhere.

A ball is thrown at speed v from the origin

What is the minimum angle to the vertical at which the ball can be thrown so that its distance to the origin is always increasing (i.e. the distance d is always increasing with time)? Give your answer to 3 significant figures.

https://isaacphysics.org/questions/b...2020_21_week12

I have the formulas for both the x and y direction of the distance from the origin. With pythagoras I can find d with these two formulas. I tried to use the discriminant: t(t^2*g^2-3v*cos(theta)*g*t+2v^2) and set it to zero but I can't get theta because I don't have any values for the velocity or time...

I would be grateful for any help.
0
reply
swinroy
Badges: 12
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 11 months ago
#2
Have a look at HINT 4
It solves it for you
the gradient function has to be greater than zero for all times t
The only way this can happen is for the discriminant of the quadratic equation in t must be greater than ZERO
This will produce an inequality expression in Cos (Theta) hence get minimum angle
You dont need to know the speed at all
1
reply
fastnfuriouseng
Badges: 5
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report Thread starter 11 months ago
#3
(Original post by swinroy)
Have a look at HINT 4
It solves it for you
the gradient function has to be greater than zero for all times t
The only way this can happen is for the discriminant of the quadratic equation in t must be greater than ZERO
This will produce an inequality expression in Cos (Theta) hence get minimum angle
You dont need to know the speed at all
Oh thank you so much! I forgot to use the discriminant
0
reply
swinroy
Badges: 12
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 11 months ago
#4
(Original post by fastnfuriouseng)
Oh thank you so much! I forgot to use the discriminant
Glad to have helped
Please hit the thumbs up button to give credit
Cheers
1
reply
YGSK
Badges: 11
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 1 month ago
#5
(Original post by swinroy)
Have a look at HINT 4
It solves it for you
the gradient function has to be greater than zero for all times t
The only way this can happen is for the discriminant of the quadratic equation in t must be greater than ZERO
This will produce an inequality expression in Cos (Theta) hence get minimum angle
You dont need to know the speed at all
Sorry i am doing the same question now, and i am so confused why the discriminant of the gradient function has to be greater than 0. Surely if the gradient function is always greater than 0, then the curve never intersects the axis, so the discriminant is always less than 0?
0
reply
mqb2766
Badges: 19
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report 1 month ago
#6
(Original post by YGSK)
Sorry i am doing the same question now, and i am so confused why the discriminant of the gradient function has to be greater than 0. Surely if the gradient function is always greater than 0, then the curve never intersects the axis, so the discriminant is always less than 0?
The minimum value will occur when the discrimiant = 0, but as you say, you want the gradient to be positive, so the quadratic must be positive so the discriminant is < 0.

As a thought experiment, take theta=0, so cos(theta)=1 and the discriminat is 9-8. > 0 This scenario obviously does not have the distance always increasing as it goes straight up, then the distance is decreasing when it comes down.
Last edited by mqb2766; 1 month ago
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What do you find most difficult about revising?

Prioritising what to revise (14)
11.02%
Remembering what was covered in lessons (2)
1.57%
Finding the time to revise (8)
6.3%
Finding the motivation (35)
27.56%
Getting distracted (17)
13.39%
Procrastination (50)
39.37%
Having the tools you need (0)
0%
Something else (tell us in the thread) (1)
0.79%

Watched Threads

View All