# Simple (was thought to be) Gravity Question

[SOLVED] Simple (was thought to be) Gravity Question

## Homework Statement

A ball is rolled up a constant slope. After 3.6 s it reaches its maximum displacement of 2.6m, and then begins to roll back down. What was the initial velocity of the ball when it started up the slope?

## The Attempt at a Solution

i know that d=vt of course, but, 2.6=v(3.6), and V would equal 2.6 divided by 3.6 which is .722, and the answer in the key is 1.4m/s?? In all likelihood, its probably something extremely obvious?

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You can't use d=vt because there is an acceleration due to gravity
use d=vt-.5at^2

you mean d=Vit+.5at^2, emphasis on plus sign?

well yes, in this case the acceleration is negative but the formula you gave is the more general form.

you mean d=Vit+.5at^2, emphasis on plus sign?
edit: thanks by the way, the question screwed me up w/ the word constant, plus im taking my very first physics course (phys. 11)----one more question is i might have quite a few questions today.....will mods get mad for me making quite a few threads? (maybe 10?)

If they're the same types of problems I'd just stick them in the same thread.