Velocity of Baseball at 47 m Height

[bradsmith]

Homework Statement

A baseball is hit directly upward with an initial speed of 37 m/s. Find the velocity of the ball when it is at a height of 47 m.

Homework Equations

F=ma
F=ag

The Attempt at a Solution

F=(37)(-9.8)=-362.6/47=7.7[/B]

 

[homo-sapiens]

It's a simple kinematics question. check your relevant equation because those equations don't look correct to me.

 

[HallsofIvy]

I don't know where you got "F= ag" but, normally, "a" is acceleration and "g" is "acceleration due to gravity. And then in you calculation below, you replace "a" with the initial speed. It certainly is NOT true that "speed times acceleration due to gravity " is equal to any common physics quantity.

In any case, there is no reason to worry about "F", force, at all. What is important is that the acceleration is -9.8 m/s^2. So the acceleration. after t seconds is -9.8t+ 37 m/s and the height reached, above the initial position, after t seconds is -4.9t^2+ 37t m. Set that equal to 47 m, solve for t, then find the velocity at that t.

 

[Ritzycat]

I also have no idea where you got those equations from.

You know acceleration due to gravity on Earth is -9.8m/s^2. Using your given values, (initial position, initial velocity, acceleration, final position), you can solve for time by using your essential kinematic equations.

 

[bradsmith]

V= √(2gh)

sqrt(37^2-2*9.8*47)=21.16

Thats what I wanted not ag and i was lazy and didnt change my F