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.
<span>So we want to know which statement is true for the body of mass m=2000kg that is lifted to a height of h=15m in t=15 s. Lets calculate each of the following: Gravity force on the body is F=m*g=2000*9.81=19620 N so a is FALSE. Potential energy of the body when it is lifted to the height of 15 m is Ep=m*g*h=2000*9.81*15=294300 J so b is FALSE. Work to lift the body is: W=Fg*h=2000*9.81*15= Ep=294300 J so c is FALSE. Power P=W/t=294300/15=19620 W So d is TRUE. </span>
Momentum - mass in motion
P=MV
P=(15,000 kg)(2.5 m/s)
P=37 500 kg x m/s to the north
Hope this helps
Explanation:
It is given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, 
Speed of light, 
Let t is the time interval required for the strong interaction to occur. The speed is given by :
So, the time interval required for the strong interaction to occur is 
. Hence, this is the required solution.
