Answer:
Explanation:
v = u +at
u = 0
a = 2.3 m /s²
t = 20 s
v = 2.3 x 20
= 46 m /s
Distance covered under acceleration of 2.3 m/s²
s = ut + 1/2 at²
= 0 + .5 x 2.3 x 20²
= 460 m
After that it moves under free fall ie g acts on it downwards .
v² = u² - 2gh , h is height moved by it under free fall
0 = 46² - 2 x 9.8 h
h = 107.96 m
Total height attained
= 460 + 107.96
= 567.96 m
b ) At its highest point ,it stops so its velocity = 0
c ) rocket's acceleration at its highest point = g = 9.8 downwards .
At highest point , it is undergoing free fall so its acceleration = g
<em>Energy</em><em> </em><em>can</em><em> </em><em>neither </em><em>be</em><em> </em><em>created </em><em>nor</em><em> </em><em>be</em><em> </em><em>destroyed</em><em> </em><em>but</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>converted</em><em> </em><em>from</em><em> </em><em>one</em><em> </em><em>form</em><em> </em><em>to</em><em> </em><em>another </em><em>.</em>
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
True
Explanation:
i searched it up and well this thing is making me do it up till 20 characters long so yea
