Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
when a element of 1 group take part in reaction, its atom looses outer electron and form positively charged ions called Cation.
Explanation:
Answer:
(a) 
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So 
(a) We know that force is given by
F = ma
So force will be 
(b) From second equation of motion we know that

We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as

(d) Time t = 1.5 sec
So 