Answer:
See below
Explanation:
Vertical position = 45 + 20 sin (30) t - 4.9 t^2
when it hits ground this = 0
0 = -4.9t^2 + 20 sin (30 ) t + 45
0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec
max height is at t= - b/2a = 10/9.8 =1.02
use this value of 't' in the equation to calculate max height = 50.1 m
it has 4.22 - 1.02 to free fall = 3.2 seconds free fall
v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL
it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s
total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s
Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m
Initial velocity of the object: 5 m/s
Explanation:
The figure in the problem is missing: find it in attachment.
The graph in the figure represents the velocity of an object (v) versus the time passed (t).
Here we are asked to find the initial velocity of the object.
This means that we have to find the velocity of the object when the time is zero, so when
t = 0
By looking at the corresponding value on the y-axis (velocity), we see that when t = 0, then
v = 5 m/s
Therefore, the initial velocity of the object is 5 m/s.
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Answer:
The horizontal displacement of the arrow is not larger than the banana split.
Explanation:
Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.
So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.
y - y₀ = ut - 1/2gt²
0 - y₀ = 0 × t - 1/2gt²
-y₀ = -1/2gt²
t² = 2y₀/g
t = √(2y₀/g)
Substituting the values of the variables, we have
t = √(2y₀/g)
= √(2 × 8848.04 m/9.8 m/s²)
= √(17696.08 m/9.8 m/s²)
= √(1805.72 s²)
= 42.5 s
The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s
So, d = vt
= 100 m/s × 42.5 s
= 4250 m
= 4.25 km
Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.
The two neutral atoms A and B have the same number of electrons and atomic number 11. So, the two elements are said to be same.
The electronic configuration of the element is the arrangement of the electrons in the atom of the element in energy levels, orbitals around the nucleus.
The electrons in the atoms of the element with lowest energy are written first before those with higher energy levels. Thus, the electronic configuration shows the electrons in the atoms of the element arranged in order of increasing energies.
The electronic configuration of atoms are given as
A = 1s² 2s² 2p⁶ 3s¹
B = 1s² 2s² 2p⁶ 5s¹
The number of electrons in both the elements is 11. Therefore, their atomic number is also the same i.e, 11. So, both the elements are the same.
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