M= ?
g=9.8 m/s (2)
h=20 m
Eg=362,600 J
Eg/mg
362,600 J/9.8 m/s (2) x 20 m
=1,850 m
Answer:
A ball is thrown straight up with a speed of 30
m/s. What is the maximum height reached by
the ball?
<span>d. electron
J J Thomson discovered the electron, and it was put in his model of the atom.</span>
Answer:
The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>
Explanation:
Given:
Upward direction is positive. So, downward direction is negative.
Tota time the ball remains in air (t) = 8.0 s
Net displacement of the ball (S) = Final position - Initial position = 0 m
Acceleration of the ball is due to gravity. So,
(Acting down)
Now, let the initial velocity be 'u' m/s.
From Newton's equation of motion, we have:
![S=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
Plug in the given values and solve for 'u'. This gives,
![0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s](https://tex.z-dn.net/?f=0%3D8u-0.5%5Ctimes%209.8%5Ctimes%208%5E2%5C%5C%5C%5C8u%3D4.9%5Ctimes%2064%5C%5C%5C%5Cu%3D%5Cfrac%7B4.9%5Ctimes%2064%7D%7B8%7D%5C%5C%5C%5Cu%3D4.9%5Ctimes%208%3D39.2%5C%20m%2Fs)
Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.