How Long Does it Take for Balls to Reach the Ground? Projectile Motion Problem

[ProBasket]

The balls are released from rest at a height of 5.0m at time t=0s. How long $$t_g$$ does it take for the balls to reach the ground? Use 10 m/s^2 for the magnitude of the acceleration due to gravity.

i'll use the position formula for this problem:

$$y = y_0 + V_{y_0}*t - 1/2*g*t^2$$

well i know that
$$y_0$$ = 5.0 m
g = 10 m/s^2

i'm missing $$V_{y_0}$$

if i plug in t = 0, wouldn't it just cancel everything out? but it doesn't make sense.

 

[mikeu]

The balls are released from rest, which tells you that their initial velocty $$V_{y_0}$$ is zero. If you plug in t=0 you just get $$y=y_0$$, or 5.0 m, which just tells you that the balls do in fact start where they start (always a good check ).

What you want to do to solve the problem is find out what value of t will give you $$y=0$$, since that's the 'height' of the ground.

 

[wais]

You are correct, if we plug in t=0, all the terms involving time will cancel out and we will be left with only the initial height (y_0). This is because at t=0, the ball has not yet started to move and so its velocity (V_{y_0}) and acceleration (g) are both zero.

To solve for the time it takes for the ball to reach the ground (t_g), we need to set the position (y) equal to zero since that is the height of the ground. So our equation becomes:

0 = 5.0 m + 0*t - 1/2*10 m/s^2 * t^2

Simplifying this further, we get:

0 = 5.0 m - 5 m/s^2 * t^2

Now we can solve for t:

t = √(5.0 m / 5 m/s^2) = 1 second

Therefore, it will take 1 second for the ball to reach the ground. This is a classic example of projectile motion, where the vertical motion of the ball is affected by gravity while the horizontal motion remains constant.