- #1
John O' Meara
- 325
- 0
The acceleration of gravity can be measured by throwing an object upward and measuring the time that it takes to pass two given points in both directions. Show that the time it takes the object to pass a horizontal line A in both directions is Ta, and the time to pass a second line B in both direction is Tb, then, if acceleration is constant, its magnitude is
g = 8h/(Ta^2 - Tb^2), where h is the height between parallel lines A and B.
V^2=U^2-2gh, V=U+gt; coming down, where U is some unknown initial velocity
V^2+U^2+2gh, V=U-gt; going up
I am a bit rusty as regards the use of the equations of motion for constant acceleration. I have h=(Vb1-Va1)/2g, where Va1 is the velocity the first time the body passes line A on the way up and Vb1 is the vel' the first time the body passes line B on the way up. But I am not sure that I have started on the correct track? I end up with 2h = 2U(Ta1 - Tb1) + g(Tb1^2 - Ta1^2). Where Ta1 is the time the body passed line A (on the way up) and Tb1 is the time the body passed line B (on the way up). It would be great if someone could stat me off. Thanks in advance.
g = 8h/(Ta^2 - Tb^2), where h is the height between parallel lines A and B.
V^2=U^2-2gh, V=U+gt; coming down, where U is some unknown initial velocity
V^2+U^2+2gh, V=U-gt; going up
I am a bit rusty as regards the use of the equations of motion for constant acceleration. I have h=(Vb1-Va1)/2g, where Va1 is the velocity the first time the body passes line A on the way up and Vb1 is the vel' the first time the body passes line B on the way up. But I am not sure that I have started on the correct track? I end up with 2h = 2U(Ta1 - Tb1) + g(Tb1^2 - Ta1^2). Where Ta1 is the time the body passed line A (on the way up) and Tb1 is the time the body passed line B (on the way up). It would be great if someone could stat me off. Thanks in advance.