What is the Speed of a Thrown Ball at its Highest Point and its Maximum Height?

[pb23me]

Homework Statement

A .4 kg ball is thrown with a speed of 12 m/s at an angle of 33 degrees. What is its speed at its highest point, and how high does it go? Use conservation of energy and ignore air resistance.

Homework Equations

mgy_i+1/2mv_i²=mgy_f+1/2mv_f²

The Attempt at a Solution

im not sure what velocity to use... in the x direction or y direction? And am i supposed to assume the ball is thrown from point zero? If i use the velocity in the y direction and assume its thrown from point zero... then v_f=0 and y_i=0 so 1/2(6.5)²=9.8(y_f) ... y_f=2.18m and v_i=sin33(12)

 

[tiny-tim]

hi pb23me!

start with the speed at its highest point …

what is that? ​

 

[pb23me]

ok I am a little confused about speed i guess. when you have an object that's moving simultaneously in two directions which is the speed? I am assuming your talking about the y direction in that case it would be zero. If your talking about the x direction then its 10.1 m/s

 

[tiny-tim]

… when you have an object that's moving simultaneously in two directions which is the speed? …

ah, you need to learn the definitions …

speed is the magnitude of the velocity

speed is a scalar (an ordinary number), and velocity is a vector

so a velocity might be written v = (3,4) or v = 3i + 4j,

or it might be given as a magnitude and a direction, ie v is 5 at an angle tan^-1(4/3)

the speed (usually written "v", not in bold font … unfortunately, sometimes velocity is also not written in bold, which is confusing) is found by using Pythagoras on the components … v = √(3² + 4²) = 5

or in this case v = √(v_x² + v_y²)

 

[pb23me]

ok cool so v=$$\sqrt{10.1^2+0}$$=10.1 m/s

 

[pb23me]

so 72=9.8(y_f)+1/2(10.1)²...y_f=2.14m

 

[tiny-tim]

(have a square-root: √ )

yup!

(except i rounded off later than you did, and got 2.18, or 2.2)

 

[pb23me]

awesome thanx, just one last question.. so when I am using the conservation of momentum equation, and conservation of energy equation do i always use the speed rather than the velocity?

 

[tiny-tim]

awesome thanx, just one last question.. so when I am using the conservation of momentum equation, and conservation of energy equation do i always use the speed rather than the velocity?

no …

energy is a scalar, like speed, so we use speed (1/2 mv² etc)

momentum is a vector, like velocity, so we use velocity (mv etc), except that we can always break it down into components in a particular direction, which is what we usually do …

eg we do conservation of momentum in the x-direction using mu_x mv_x etc, and maybe then also conservation of momentum in the y-direction using mu_y mv_y etc