For a flourish at the end of her act, a juggler tosses a single ball high in the air. She catches the ball 3.6 s later at the same height from which it was thrown.
Part A-What was the initial upward speed of the ball?2sf
if the weather channel guessed that it would rain, you would not believe them even if they have evidence of a thunder storm. My sister said that she assumed that the ball's speed would be the same going up and down so she gave "evidence" for why she thought the answer was 1.8
reaches top 3.6/2 or 1.8 seconds after launch
at top, v = 0
but
v = Vi - g t
in meters, seconds g = 9.81 m/s^2
so
0 = Vi -9.81 t
0 = Vi -9.81*1.8
Vi = 17.7 meters/second
9.81 m/s is the constant for gravity, or acceleration during free fall. He is correct and his math works out as well.
Because you were only given the time it took for the ball to reach its initial position half of the time it spent in travel is against gravity and the other in free fall.
So the formula I used to get the answer was: Vi=(1/2t)*gravity(9.81m/s)
Hope this helped.
How did Damon get 9.81 meters/second?
To find the initial upward speed of the ball, we can use the following equation of motion:
v_f = v_i + a*t
Where:
v_f = final velocity (which is 0 since the ball reaches its highest point and then falls back down)
v_i = initial velocity (what we want to find)
a = acceleration (in this case, due to gravity, it is -9.8 m/s^2)
t = time taken (which is 3.6 s)
Since the final velocity is 0, we can rearrange the equation to solve for the initial velocity (v_i):
v_i = -(a * t)
Substituting the values:
v_i = -(9.8 m/s^2 * 3.6 s)
= -35.28 m/s
Since the upward direction is considered positive, we take the magnitude of the velocity:
v_i = 35.28 m/s (rounded to 2 significant figures)
Therefore, the initial upward speed of the ball is approximately 35.28 m/s.
I think that the answer is 1.8
3.6/2(for the time it toke to go up then back down)=1.8
I am only in 7th grade so i am just guessing hope this is the right answer though :)