We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
Answer:
![[F]=[MLT^{-2}]](https://tex.z-dn.net/?f=%5BF%5D%3D%5BMLT%5E%7B-2%7D%5D)
Explanation:
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :
F = m × a
F is the applied force
m is the mass of the object
a is the acceleration due to gravity
We need to find the dimensions of force. The dimension of force m and a are as follows :
![[m]=[M]](https://tex.z-dn.net/?f=%5Bm%5D%3D%5BM%5D)
![[a]=[LT^{-2}]](https://tex.z-dn.net/?f=%5Ba%5D%3D%5BLT%5E%7B-2%7D%5D)
So, the dimension of force F is, . Hence, this is the required solution.
We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times
Answer:
C. Equals
Explanation:
Law of reflection Equals the angle of incidence
i do not have an answer because it depends on the size and the distance lol
