Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is
The values are
We must find the values of t such that the height of the camera is 0 (when it hits the ground)
Multiplying by 2
Clearing the coefficient of 
Plugging in the given values, we reach to a second-degree equation
The equation has two roots, but we only keep the positive root
Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is
I’m assuming D because it’s the most reasonable
I'm probably going to have to say C. E as it seems the steepest right around there. If I'm wrong on that, it has to be B. B
This question can be solved by using the equations of motion.
a) The initial speed of the arrow is was "9.81 m/s".
b) It took the arrow "1.13 s" to reach a height of 17.5 m.
a)
We will use the second equation of motion to find out the initial speed of the arrow.
where,
vi = initial speed = ?
h = height = 35 m
t = time interval = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
<u>vi = 9.81 m/s</u>
b)
To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 17.5 m
vi = initial speed = 9.81 m/s
t = time = ?
Therefore,
solving this quadratic equation using the quadratic formula, we get:
t = -3.13 s (OR) t = 1.13 s
Since time can not have a negative value.
Therefore,
<u>t = 1.13 s</u>
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
This is correct, I just did the test. Yes, displacement is 45 meters, elapsed time is three seconds, and the direction is toward the goal.
