Projectile Motion Kinematics: Finding maximum height

  • #1

Homework Statement


A ball is kicked with an initial speed of 20m/s at an angle of 50° above the horizontal. What maximum height does the ball reach?

Homework Equations


  • Vfx=VicosΘ
  • Δx=VicosΘt
  • Vfy=(VisinΘ)+ay⋅t
  • (Vfy)^2=(VisinΘ)^2+2ay⋅Δy
  • Δy=1/2(Vfy+(VisinΘ))⋅t

The Attempt at a Solution


I know that the final velocity for both x and y is 20m/s if that is the case. The acceleration for the x-direction is 0m/s^2 while its -9.8m/s^2 due to earth gravitational acceleration. Vy at max height is 0.

Known
  • Vfx and Vfy= 20m/s
  • Θ=50°
  • Vfy=0m/s
  • ax=0m/s^2
  • ay=-9.8m/s^2
  • T=?
  • Δy=?
  • Δx=?

I'm assuming that the question is asking for max height as Δy in the y-direction. I was trying to solve for time in order to use the equation to solve for Δy but I can't see how to do that. I'm not sure if there is another way that I could do this or if I just labeled things incorrectly.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


A ball is kicked with an initial speed of 20m/s at an angle of 50° above the horizontal. What maximum height does the ball reach?

Homework Equations


  • Vfx=VicosΘ
  • Δx=VicosΘt
  • Vfy=(VisinΘ)+ay⋅t
  • (Vfy)^2=(VisinΘ)^2+2ay⋅Δy
  • Δy=1/2(Vfy+(VisinΘ))⋅t

The Attempt at a Solution


I know that the final velocity for both x and y is 20m/s if that is the case. The acceleration for the x-direction is 0m/s^2 while its -9.8m/s^2 due to earth gravitational acceleration. Vy at max height is 0.

Known
  • Vfx and Vfy= 20m/s
  • Θ=50°
  • Vfy=0m/s
  • ax=0m/s^2
  • ay=-9.8m/s^2
  • T=?
  • Δy=?
  • Δx=?

I'm assuming that the question is asking for max height as Δy in the y-direction. I was trying to solve for time in order to use the equation to solve for Δy but I can't see how to do that. I'm not sure if there is another way that I could do this or if I just labeled things incorrectly.
You do not need to "assume" anything about what the question is asking---it tells you exactly what it wants.

Anyway, the initial speed is ##v_0 = 20## m/s, but at an angle of 50° up. From that, you need to figure out the x- and y-components ##v_{0x}## and ##v_{0y}##. They are NOT 20 m/s: draw a picture to see why.

After you have found ##v_{0x}## and ##v_{0y}## the rest of the question ought to be fairly straightforward..
 
  • #3
I found the x and y components. im a bit confused now..
 
  • #4
bobob
Gold Member
165
97
Why are you using an equation to find the time? You have a much more useful equation in terms of squared initial and final velocities, acceleration and distance: V_f^2 = V_i^2 + 2ad. You wrote down the expression for the initial vertical velocity and you said you know the final vertical velocity is zero at the peak of its flight. You know the acceleration.
 
  • #5
Why are you using an equation to find the time? You have a much more useful equation in terms of squared initial and final velocities, acceleration and distance: V_f^2 = V_i^2 + 2ad. You wrote down the expression for the initial vertical velocity and you said you know the final vertical velocity is zero at the peak of its flight. You know the acceleration.
i tried doing it that way but i didnt get the answer. i dont know if i labeled things correctly.
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,533
6,430
I found the x and y components. im a bit confused now..
So what did you get for the y component?
 
  • #8
is this correct. I solved for t using
  • Vfy=(VisinΘ)+ay⋅t
i got 1.56s

I then used

  • Δy=1/2(Vfy+(VisinΘ))⋅t
and got 11.9m or 12m
 
  • #9
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
That looks correct to me, however you could do it in one step if you used another of the SUVAT equations...

V2 = U2 + 2as
 
  • #10
That looks correct to me, however you could do it in one step if you used another of the SUVAT equations...

V2 = U2 + 2as
I see :| my teacher gave us like 8 equations to memorize... I have no clue of any of the other methods
 
  • #11
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,533
6,430
I see :| my teacher gave us like 8 equations to memorize... I have no clue of any of the other methods
The SUVAT equations for linear motion under constant acceleration are named after the five variables: s=distance, u=initial speed, v=final speed, a=acceleration, t=time.
Any three will determne the remaining two, so there are five equations, each omitting one variable.
The trick, then, is to identify the four variables of interest and use the equation involving those.
In the present case it was s, u, v and a. Note that this particular equation is equivalent to the work conservation equation for constant force: ΔKE=F.d.
 
  • #12
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
Note that this particular equation is equivalent to the work conservation equation for constant force: ΔKE=F.d.
If you are also studying physics and have covered conservation of energy then this is how you get there...

Conservation of energy says..
Final Kinetic Energy = Initial Kinetic Energy + work done on object
0.5mV2 = 0.5mU2 + ma*s

multiply both sides by 2
mV2 = mU2 + m2as

mass cancels..
V2 = U2 + 2as
 
  • #13
BvU
Science Advisor
Homework Helper
14,106
3,529
my teacher gave us like 8 equations to memorize
Thing to do is understand what's going on. That way you don't need to memorize so much, or, indeed, nothing at all.
 
  • #14
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
+1

There are ways to understand them, for example..

S = (U+V)t/2

...is the average velocity (V+U)/2, multiplied by time t.

You can also think of it as the area under a graph of velocity vs time.

Velocity vs time.png
 

Attachments

  • #15
PeterO
Homework Helper
2,426
48

Homework Statement


A ball is kicked with an initial speed of 20m/s at an angle of 50° above the horizontal. What maximum height does the ball reach?

Homework Equations


  • Vfx=VicosΘ
  • Δx=VicosΘt
  • Vfy=(VisinΘ)+ay⋅t
  • (Vfy)^2=(VisinΘ)^2+2ay⋅Δy
  • Δy=1/2(Vfy+(VisinΘ))⋅t

The Attempt at a Solution


I know that the final velocity for both x and y is 20m/s if that is the case. The acceleration for the x-direction is 0m/s^2 while its -9.8m/s^2 due to earth gravitational acceleration. Vy at max height is 0.

Known
  • Vfx and Vfy= 20m/s
  • Θ=50°
  • Vfy=0m/s
  • ax=0m/s^2
  • ay=-9.8m/s^2
  • T=?
  • Δy=?
  • Δx=?

I'm assuming that the question is asking for max height as Δy in the y-direction. I was trying to solve for time in order to use the equation to solve for Δy but I can't see how to do that. I'm not sure if there is another way that I could do this or if I just labeled things incorrectly.
This is fundamentally the same as your "Stone thrown from a bridge" question you posted. Solve one of them and you should see how to solve them all. You need to do some reading up on projectile motion.
 
  • Like
Likes davenn

Related Threads on Projectile Motion Kinematics: Finding maximum height

Replies
8
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
7
Views
79K
Replies
5
Views
3K
  • Last Post
Replies
2
Views
850
Replies
9
Views
3K
Replies
13
Views
328
Replies
1
Views
2K
  • Last Post
Replies
3
Views
4K
Top