Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4.
The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
Answer:
g' = 13.5 m/s²
Explanation:
The acceleration due to gravity on surface of earth is given by the formula:
g = GMe/Re² --------------- euation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
Me = Mass of Earth
Re = Radius of Earth
Now, the the acceleration due to gravity on the surface of Kepler-62e is:
g' = GM'/R'² --------------- euation 1
where,
g' = acceleration due to gravity on surface of Kepler-62e
G = Universal Gravitational Constant
M' = Mass of Kepler-62e = 3.57 Me
R' = Radius of Kepler-62e = 1.61 Re
Therefore,
g' = G(3.57 Me)/(1.61 Re)²
g' = 1.38 GMe/Re²
using equation 1:
g' = 1.38 g
where,
g = 9.8 m/s²
Therefore,
g' = 1.38(9.8 m/s²)
<u>g' = 13.5 m/s²</u>
Answer:
A 1.5 metres per second squared B 0 C 6 0.5
b) 9m
Explanation:
a= v-a/t
a= acceleration measured in metres per second squared
v= final speed measured in metres per second
u= initial speed measured in metres per second
t= time measured in second
A
a=v-a/t
a= 3-0/2
a= 1.5 metres per second squared
Distance in a velocity time graph = area under
To calculate the distance in the first 4 seconds
You divide the graph into triangles and rectangles
Area A (triangle )= 1/2 × base × height
=1÷2 ×2×3
= 3m
Area B (rectangle)= L×B
=2×3
=6m
Area A + Area B= 3+6
= 9m
Answer:
heavy box with no heels on it
