A coin is launched horizontally with a speed of 7.0 m/s from the balcony of a fifth-storey apartment. Determine the horizontal velocity of the coin 1.5 s after it is launched.
Question 1 options:
a) 10.5 m/s [forward]
b) 7.0 m/s [forward]
c) 0.0 m/s
d) 4.67 m/s [forward]
e) 7.0 m/s [down]
11 answers
The horizontal velocity of the coin remains constant when it is launched since there are no horizontal forces acting on it (assuming air resistance is negligible).
Given that the initial horizontal speed is 7.0 m/s and there are no horizontal forces acting on it, the horizontal velocity after 1.5 seconds will still be:
7.0 m/s [forward]
So the correct answer is:
b) 7.0 m/s [forward]
To find the maximum height of the rocket, we can use the following kinematic equation for vertical motion:
\[ v_y^2 = v_{0y}^2 - 2g h \]
Where:
First, we need to calculate the initial vertical velocity \( v_{0y} \):
\[ v_{0y} = v \sin(\theta) \]
Where:
Calculating \( v_{0y} \):
\[ v_{0y} = 26 \sin(15°) \approx 26 \times 0.2588 \approx 6.728 , \text{m/s} \]
Now we can substitute \( v_y = 0 \), \( v_{0y} \approx 6.728 , \text{m/s} \), and \( g = 9.81 , \text{m/s}^2 \) into the kinematic equation:
\[ 0 = (6.728)^2 - 2(9.81)h \]
Solving for \( h \):
\[ 0 = 45.229 - 19.62h \]
\[ 19.62h = 45.229 \]
\[ h = \frac{45.229}{19.62} \approx 2.31 , \text{m} \]
Thus, the maximum height the rocket reaches is approximately \( 2.3 , \text{m} \).
The correct answer is:
c) 2.3 m
When considering the acceleration of the air-powered rocket after launch, we need to account for the forces acting on it.
Gravity is acting downward with a constant acceleration of approximately \( 9.81 , \text{m/s}^2 \).
If we assume that the rocket is powered by air propulsion, there may be an upward acceleration during the powered flight (while the engines are firing), but once the rocket is no longer powered, the only acceleration acting on it is the acceleration due to gravity, which will be \( 9.81 , \text{m/s}^2 \) in the downward direction.
Since the problem does not specify whether the propulsion is active, we'll assume that you're asking for the acceleration due to gravity once the rocket is launched. Therefore, the acceleration of the rocket after launch (due to gravity) is:
9.8 m/s² [down]
So, the correct answer is:
a) 9.8 m/s² [down]
When the coin is launched horizontally, its horizontal motion is independent of its vertical motion. The only force acting on the coin in the vertical direction is gravity, which causes it to accelerate downward at approximately \( 9.81 , \text{m/s}^2 \).
Regardless of the time elapsed or the horizontal velocity, the acceleration due to gravity remains constant. Thus, 1.5 seconds after it is launched, the acceleration of the coin is still:
9.8 m/s² [down]
The correct answer is:
c) 9.8 m/s² [down]
To determine the swimmer's velocity relative to the shore, we need to combine the swimmer's velocity relative to the water and the velocity of the water current.
We can represent these two velocities as vectors:
To find the resultant velocity of the swimmer relative to the shore, we can use the Pythagorean theorem to calculate the magnitude and inverse tangent to find the angle.
Using the Pythagorean theorem:
\[ v = \sqrt{(v_N^2 + v_W^2)} \]
where:
\[ v = \sqrt{(1.0^2 + 1.5^2)} = \sqrt{(1 + 2.25)} = \sqrt{3.25} \approx 1.80 , \text{km/h} \]
Using the inverse tangent to find the direction:
\[ \theta = \tan^{-1}\left(\frac{v_W}{v_N}\right) = \tan^{-1}\left(\frac{1.5}{1.0}\right) = \tan^{-1}(1.5) \approx 56.31° \text{ (from the north to the west)} \]
Thus, the swimmer's velocity relative to the shore is approximately:
\[ \text{Magnitude} \approx 1.80 , \text{km/h}, \quad \text{Direction} \approx 56° \text{ W of N} \]
So, the correct answer is:
c) 1.80 km/h [N 56° W]
To determine the vertical velocity of the coin 1.5 seconds after it is launched, we need to consider the effect of gravity on the coin.
When launched horizontally, the vertical motion of the coin is affected by the acceleration due to gravity, which is approximately \(9.81 , \text{m/s}^2\) downward.
The vertical velocity (\(v_y\)) after a time (\(t\)) can be calculated using the following formula:
\[ v_y = g \cdot t \]
Where:
Substituting the values:
\[ v_y = 9.81 , \text{m/s}^2 \cdot 1.5 , \text{s} = 14.715 , \text{m/s} \]
Since the coin is falling downwards, we indicate this direction as negative. Thus:
\[ v_y \approx 14.7 , \text{m/s} [\text{down}] \]
So, the correct answer is:
d) 14.7 m/s [down]
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