Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
Imma guess A! Idk if it’s 100% correct tho so I’d check that!
Answer:
have a component along the direction of motion that remains perpendicular to the direction of motion
Explanation:
In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.
circular movement
F = m a
a = v² / r
F = m v²/R
where the force is perpendicular to the velocity, all the force is used to change the direction of the velocity
in linear motion
F = m a
where the force is parallel to the acceleration of the body, the total force is used to change the modulus of the velocity
the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion
No, because sometimes you have to stop at stop signs and stop lights.
This situation has a basis such that the solid sphere and the hoop has the same mass. The analysis could be made<span> backwards . The ball will decelerate fastest, so not roll as high. The sphere will accelerate faster, but this also means it decelerates faster on the way up. Hence the answer is the hoop if the masses are equal </span>
