For this use the formula:
d = Vo * t - (at^2) / 2
Clearing t:
t = d/(v + 0.5*a)
Replacing:
t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)
Resolving:
t = 5 m / (7.2 m/s + (-0.55 m/s²)
t = 5 m / 6.65 m/s
t = 0.75 s
Result:
The time will be <u>0.75 seconds.</u>
Answer:
Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height
Explanation:
As we know that initially ball is held at height h = 40 cm
So here we can say that kinetic energy of the ball is zero and potential energy is given as

now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies
So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height
While if we assume that there is no energy loss during collision then in that case ball will reach to same height again
Answer:
a = 2.5 m/s^2
Explanation:
u = 0
v = 25
t = 10
(using first eq. of motion)
a = (v - u) /t
a = (25 - 0) /10
a = 25/10
a = 2.5 m/s^2
Answer:
true
Explanation:
I believe the answer i chose because the earths atmosphere is mostly made up of different things that causes the earth to interact with human life and also interacts with what's in the atmosphere like energy oxygen carbon dioxide and all the stuff like that i hope its right.