Maximum Height Reached by Slingshot-Launched Apple?

[hmlaa]

Hello! My question is...What is the maximum height obtained by a 125 g apple that is slung from a slingshot at an angle of 78 degrees from the horizontal with an initial velocity of 18 m/s?

Thanks!

 

[Fightfish]

Any attempts/ideas on your part? This has to do with conservation of energy.

 

[hmlaa]

I thought I could use .5Vo^2/g but I don't know how to take mass into account =\

 

[estalas]

In an ideal situation without air resistance ect. all mass undergoes same downward acceleration of g near the surface of the earth. Hence, mass is not used in this question.

Start by finding the vertical component of the apple velocity

 

[APHdotCom]

hello! My question is...what is the maximum height obtained by a 125 g apple that is slung from a slingshot at an angle of 78 degrees from the horizontal with an initial velocity of 18 m/s?

Thanks!

E1 = E2

The apple is still moving at the top of the arc, so there's still some kinetic energy left over.

KE1 = KE2 + PE2

(1/2) * m * (v1)^2 = (1/2)*m*(v2)^2 + m*g*h2

At the peak of the arc, there is no y-component of velocity. All that's left at the top is the x-component. Since there is no acceleration in the x direction, the x-velocity of the apple at the top will be the same as it was at the beginning

vx1 = vx2 = v2 = v1*cos(78)


Plug v2 into energy equation.

(1/2) * m * (v1)^2 = (1/2)*m*(v1*cos(78))^2 + m*g*h2

Optional step: factor "m" out of every term. The "m"s drop and you're just left with...
(1/2)*(v1)^2 = (1/2)*(v1*cos(78))^2 + g*h2


Solve for h2

h2 = [ (1/2)*m*(v1)^2 - (1/2)*m*(v1*cos(78))^2 ] / (m*g)

= ((1/2)*0.125kg*(18m/s)^2 - (1/2)*0.125kg*(18m/s*cos(78degrees))^2)/(0.125kg*9.80665m/s^2)


h2 = 15.805314 meters

Hope this helps!

-Alex

http://www.alexpleasehelp.com

 

[hmlaa]

Thank y9u all!