Magnitude of change is 15km/h
change in direction is to the right?
Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
Answer:
In kinematics questions we need to separate the question into different parts if the acceleration changes. Here, there are three time intervals where acceleration is different.
1) a(t) = 96t. We can find the velocity function of the rocket by integrating the acceleration function. Then we can integrate again to find the position function.
'C' is the integration constant. We can find this constant by investigating the initial conditions.
We know that the rocket is initially at rest, so 'C' should be zero.
Again, the rocket started from ground zero, so C = 0.
We should conclude the first part by calculating the final position and final velocity of the rocket.
2) For the second part, the rocket is in free fall, so
The maximum height that the rocket reaches is when its velocity is zero.
So,
The maximum height is
The final positions for the part 2 is
3) With the parachute, the velocity is dropped from -276.4 to 16 in 5 s.
The rocket lands
Answer:
<em>61.9°</em>
Explanation:
The formula for calculating the workdone is expressed as
Workdone = Fdsin theta
F is the force applied on the crate
d is the distance covered
theta is the angle that the rope makes with the horizontal
Given
F = 40N
d = 7m
Workdone = 247J
Substituting into the formula:
247 = 40(7)sin theta
247 = 280sin theta
sin theta = 247/280
sin theta = 0.88214
theta = arcsin(0.88214)
theta = 61.9°
<em>Hence the angle that the rope makes with the horizontal is 61.9°</em>