The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
You should put when you will leave, where you will be, and what time you will get back.
It’s A because it stays in motion whenever you drop it
Answer:

The Magnitude of electric field is in the upward direction as shown directly towards the charge
.
Explanation:
Given:
- side of a square,

- charge on one corner of the square,

- charge on the remaining 3 corners of the square,

<u>Distance of the center from each corners</u>


∴Distance of center from corners, 
Now, electric field due to charges is given as:

<u>For charge
we have the field lines emerging out of the charge since it is positively charged:</u>

<u>Force by each of the charges at the remaining corners:</u>

<u> Now, net electric field in the vertical direction:</u>


<u>Now, net electric field in the horizontal direction:</u>


So the Magnitude of electric field is in the upward direction as shown directly towards the charge
.