<h2>
Answer: 12 s</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.
In this sense, the main movement equation in the Y axis is:
(1)
Where:
is the instrument's final position
is the instrument's initial position
is the instrument's initial velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of planet X.
As we know and 
when the object hits the ground, equation (1) is rewritten as:
(2)
Finding :
(3)
(4)
(5)
Finally:
Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
Y, bc the height of the bounce back is higher than x
Answer:
Vprom = 0.00347[km/min]
Explanation:
We can calculate each of the average speeds and then perform the overall average between the two speeds.
V1 = 6/54
V1 = 0.111[km/min]
V2 = 1/16
V2 = 0.0625[km/min]
![V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]](https://tex.z-dn.net/?f=V_%7Bprom%7D%20%3D%20%5Cfrac%7BV_%7B1%7D%20%2B%20V_%7B2%7D%7D%7B2%7D%20%20%5C%5CV_%7Bprom%7D%20%3D%20%5Cfrac%7B0.1111%20%2B%200.0625%7D%7B2%7D%5C%5CV_%7Bprom%7D%20%3D%200.00347%20%5Bkm%2Fmin%5D)
