<span>(a)
Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get:
18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component.
(b)
To much the same end do we derive the vertical component:
18.0*sin37.5 = v_y = 10.96 ms^-1
Which we then divide by acceleration, a_y, to derive the time till maximal displacement,
10.96/9.8 = 1.12 s
Finally, doubling this value should yield the particle's total time with r_y > 0
<span>2.24 s
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V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
Answer: The change in velocity is 20mph
Explanation: The change in velocity is the difference between the final velocity and the initial velocity.
The initial velocity is 0 and the final velocity is 20mph.
Using the formula dV=Vf-Vi
dV=20-0
dV=20mph North
<span>The weight of the spacecraft keeps changing.
</span>
<span>The mass of the spacecraft remains the same.
These are the correct answers</span>