1
$\begingroup$

Question: A ball is thrown straight up, from $3\rm m$ above the ground, with a velocity of $14\rm m/s$. When does it hit the ground?

In this question, to find the total height first, we must also take into consideration the acceleration due to gravity as well. To find the distance by which acceleration due to gravity pulls the ball down must know the Initial velocity.

In the answer key, it says that gravity changes the ball's height by a distance of $-5t^2$

Is initial velocity in this question $0$ or $14\rm m/s$? Because if it is $14$, then gravity should change the balls position by $-(14+5t^2)$, not $-5t^2$, because if I remember correctly, the formula for finding distance is $vt(\mathrm{initial velocity})+(1/2)at^2$, and not just $(1/2)at^2$

$\endgroup$
4
  • $\begingroup$ What is the hypothesis? $\endgroup$
    – Mr. Xcoder
    May 20, 2017 at 12:24
  • $\begingroup$ There is no contradiction between your recollection and the answer key. Your recollection is correct (if the ball starts from the origin). The answer key says that the effect of gravity is to change the ball's position according to $\frac{1}{2}at^2$. The other term in the formula is due to the initial velocity and has nothing to do with gravity. $\endgroup$ May 20, 2017 at 15:00
  • $\begingroup$ But we have to find the total distance by which gravity pulls down the ball, don't we? To do so we have to use the formula vt(initial velocity)+(1/2)at^2 so initial velocity is considered. $\endgroup$
    – user34782
    May 20, 2017 at 15:06
  • $\begingroup$ The ball's vertical motion has a constant acceleration, so the one-dimensional vector equation $$s=ut+\frac12at^2$$ applies. Assigning the initial position as the origin, and upwards as the positive direction, we have: constant acceleration $a = -10m/s^2,$ initial velocity $u = 14m/s,$ displacement $s = -3m.$ So, dropping the units, $$-3=14t+\frac12(-10)t^2\\t=3.$$ $\endgroup$
    – ryang
    Jul 28, 2021 at 6:49

1 Answer 1

0
$\begingroup$

You've ignored the multiplication of initial velocity by time of flight. The total distance above ground is given by $$3+14t-5t^2$$which then must be solved for equality to zero distance above ground, and account for common sense.

Gravitational acceleration creates a change in distance of $$-5t^2$$ if we use 1 significant figure. The only thing affecting the gravitational changes is gravity itself, which has no dependence on velocity (at least in his scale).

The initial velocity changes distance by a separate factor,$$v_it$$ which does not depend on gravity, hence these terms are separate in the equation. The constant term is straightforwardly the height at time zero.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .