An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50m away at a constant speed of 2.50m/s, returning just in time to catch the falling ball. (a)With what minimum initial speed must she throw the ball upward to accomplish this feat? (b)How high above its initial position is the ball just as she reaches the table?
i know just v1=25.7
bat the secound part i don't know about it ..
What you are proably doing wrong is that you a taking the Dis[lacement as it comes.Be careful because the ball could be on its way back down. i wont do the question out but you have to find it max heigth by entering the final velocity as 0 and then subtract this answer for the one you got for the displacement
Also be careful of sign if the ball is going up the acceleration is -9.81
To solve this problem, we can break it down into two parts. First, let's calculate the time taken for the entertainer to run to and from the table. Then, we can determine the minimum initial speed required to throw the ball upward to catch it in time. Finally, using the time the ball is in the air, we can calculate the height it reaches.
Let's begin with part (a), where we calculate the minimum initial speed required:
1. Calculate the time taken to run to and from the table:
Distance = 2 * 5.50m = 11.00m (running to the table and returning)
Speed = 2.50m/s
Time = Distance / Speed = 11.00m / 2.50m/s = 4.40s
2. Once we have the time taken to run to and from the table, we can calculate the time the ball is in the air. The total time will be twice the time taken by the entertainer to complete her run: 2 * 4.40s = 8.80s
3. To find the minimum initial speed required to throw the ball upward, we'll use the equation of motion for vertical motion:
v = u + at
Where:
v = final velocity (when the ball returns to the entertainer's hand, its velocity is 0)
u = initial velocity (what we need to find)
a = acceleration (in this case, acceleration due to gravity, which is -9.8m/s^2)
t = time (8.80s)
Rearranging the equation:
u = (v - at)
Since the final velocity is 0, the equation becomes:
u = -at
Substitute the known values:
u = -(9.8m/s^2)(8.80s) = -86.24m/s
The minimum initial speed required to throw the ball upward is 86.24m/s (upward because we chose the negative sign to indicate the opposite direction to gravitational acceleration).
Moving on to part (b), where we calculate the height above its initial position that the ball reaches:
1. We know the time the ball is in the air is 8.80s.
2. We can use the equation of motion for vertical motion to calculate the displacement (height) achieved by the ball:
s = ut + (1/2)at^2
Where:
s = displacement (what we need to find)
u = initial velocity (86.24m/s)
a = acceleration (in this case, -9.8m/s^2)
t = time (8.80s)
Substituting the known values:
s = (86.24m/s)(8.80s) + (1/2)(-9.8m/s^2)(8.80s)^2
s = 759.87m
The ball reaches a height of 759.87m above its initial position just as the entertainer reaches the table.