Projectile Motion: Kinematics: Finding maximum height

[TheDragon]

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?

Equations:

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

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. Vfy 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.

[Ghideon]

I'll try to give some hints by asking questions.

What is [math]V_{i} [/math] ?

7 hours ago, TheDragon said:

initial speed of 20m/s at an angle of 50°

 

7 hours ago, TheDragon said:

the final velocity for both x and y is 20m/s

If you draw a picture, how are the final velocities related to the inital speed?

 

 

Edited September 7, 2018 by Ghideon

[studiot]

Let me say that your method of writing down all the equations and all the information or question marks as a list is a good one.

But I don't know why you are using deltas for x and y  but not for time. They are not needed.

I am guessing that you have not done enough calculus to attack this problem that way? however the key to this is to realise that the horizontal acceleration is zero.

Therefore the horizontal velocity is constant.

You have this in one of your equations.

 

Therefore when the ball is halfway in time it is also halfway in horizontal distance and vice versa.

Can you use this observation to make progress?

You will need to work several of your equations together to get the desired formula.

I suggest you rewrite your equations (without the deltas) as equations of the form t = and compare them at known points in the horizontal and vertical paths.

Edited September 7, 2018 by studiot

[swansont]

13 hours ago, TheDragon said:

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?

...

I know that the final velocity for both x and y is 20m/s if that is the case.

How do you "know" this?

[Country Boy]

How you would approach this depends upon what you know.  If you know a little Calculus then you can start from the fact that there is no horizontal acceleration while the vertical acceleration is that of gravity, -g or approximately -9.8 m/s.

That  horizontal acceleration is the derivative of the horizontal velocity function: dvx/dt= 0 so vx is a constant.  You are told that, initially, the speed is  "20m/s at an angle of 50° above the horizontal" so that constant is 20 sin(50).  The horizontal velocity is the derivative of the horizontal position,  x, so the horizontal position, taking the point at which the ball is kicked to be x= 0, is x(t)= 20 sin(50)t.

The vertical acceleration is the derivative of the vertical velocity function: dvy/dt= -g so vy= -gt plus the initial vertical velocity which is 20 cos(50).  vy= -gt+ 20 cos(50).  The vertical velocity is the derivative of the vertical position, y, so taking the point at which the ball is kicked to be y= 0, y(t)= -(g/2)t^2+ 20 cos(50)t.

If you have not learned any Calculus, you probably have just learned that "x(t)= vx(0)t+ x(0)" and "y(t)= -(g/2)t^2+ vy(0)t+ y0" where vx(0), vy(0), x(0), and y(0) are the initial values put in above.  

The question asks for the highest point the ball reaches. Again, there are several ways of doing that.  If you know a little Calculus,  you know that a maximum of a differentiable function is reached where the y derivative is 0.  If not the you probably realize that, at the highest point, the vertical velocity, vy, is 0.  Or, since the vertical position function is quadratic, its graph is a parabola with vertex at its highest point- you can simply "complete the square" in the quadratic formula for the height. 

[Ghideon]

On 2018-09-07 at 1:43 AM, TheDragon said:

What maximum height does the ball reach?

Did the any of the replies guide you to the answer? Was any of the responses particularly helpful or not helpful? Some feedback would be nice.