a bowling ball is thrown down from the top of a building at 20 m/s. Find the distance the ball travels in

A) 1 s
B) 3 s

should I make the 20 m/s negative? I'm assuming the ball is just being thrown stright down...

I'm using the formula d=V(initial)*t+1/2a*t^2

I'm getting -24.9 for a and -104.1 for b if I assume that 20 m/s is negative. I just really need someone to clear this up for me. please and thank you.

Please, please, I really need gelp for this.

yes, make the intial velocity negative, and a is negative also.

d= vi*t+1/2 g t^2 in the first second.
d=-20*1 -4.9 *1=-24.9 meters, if + is upward.

oh, okay, hank you so much!

To solve this problem, let's use the formula for the distance traveled by an object under constant acceleration:

d = v(initial) * t + 1/2 * a * t^2

In this case, the initial velocity (v(initial)) is 20 m/s, the time (t) is given as 1 second for part A and 3 seconds for part B. The acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s^2.

For Part A:

Using the formula, we have:
d = 20 m/s * 1 s + 1/2 * (-9.8 m/s^2) * (1 s)^2
d = 20 m + 1/2 * (-9.8 m/s^2) * 1 m
d = 20 m - 4.9 m
d = 15.1 m

Therefore, the ball travels a distance of 15.1 meters in 1 second.

For Part B:

Using the formula, we have:
d = 20 m/s * 3 s + 1/2 * (-9.8 m/s^2) * (3 s)^2
d = 60 m + 1/2 * (-9.8 m/s^2) * 9 m
d = 60 m - 44.1 m
d = 15.9 m

Therefore, the ball travels a distance of 15.9 meters in 3 seconds.

In this problem, we assume the positive direction to be upward, so the initial velocity of 20 m/s should be positive. The negative sign is used to indicate that the acceleration is in the downward direction due to gravity. Hence, you should not make the initial velocity negative.