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.
The object will continue moving in a straight line at constant speed.
Definitely D. The brakes on a bike rub against the wheel. Not sure about the others.
Answer: The first one
Explanation: I think it's the first one because it says what is the "least" gravitational potential energy story between the prairie dog and Earth that said resting in its borrow is using less energy
Answer:
117.72kW
Explanation:
Given data
Mass m= 50kg
height x = 2m
time taken = 2 minutes= 129 seconds
let us find the work done
WD= force * distance
WD= mgx
WD= 50*9.81*2
WD= 981 Joules
Let us find the power
Power= work * time
Power= 981*120
Power= 117720
Power= 117.72 kW
Hence the power spent is 117.72kW
