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
Answer: 65.25 J
Explanation:
Kinetic Energy K.E. = 1/2 * m * v^2 ; where m is the mass of the body and v is the velocity of the body ; K.E. = 1/2 * 0.145 * 30 * 30 = 65.25 Joules
Answer:a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.
Hope This helps!!
The magnitude of the electric force between two obejcts with charge
and
is given by Coulomb's law:
where
is the Coulomb's constant
and r is the distance between the two objects.
In our problem, the distance is
, while the magnitudes of the two charges are
(we can neglect the sign of the second charge, since we are interested only in the magnitude of the force).
So, using the formula and the data of the problem, we find
To solve this we can use one of the main four kinematic equations, preferably we want something that we can use directly without solving for other values.
So in this problem we know that:
Inital velocity or Vi = 0 (started from rest)
The acceleration of the object = 7.5
The distance the object travels Δx = 87m
So, our best equation here with what we have would be:
Vf² = Vi² + 2aΔx
Now we now that Vi is 0 so:
Vf² = 2aΔx
We need final velocity so lets take the square root of Vf to isolate it:
Vf = √2aΔx
Now we plug in
Vf = √2(7.5)(87)
Vf = 36.124 m/s or 36.1 m/s
