Newton's<span> first </span>law of motion<span> has been frequently stated throughout this lesson. An</span>object<span> at rest stays at rest and an </span>object<span> in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless </span>acted<span> upon by an </span>unbalanced force<span>.</span>
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.
-- heat the magnet red-hot in a flame
-- drop it on the floor several times, or beat it with a hammer
-- place it in a strong, rapidly alternating external magnetic field
Answer:
0.5 , 54.5
Explanation:
for acceleration we should derivate the equation 2 times
x=3t³+t²/4
v=9t²+t/2
a=18t+1/2
a(0)=0.5
a(3)=54.5