If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
1 answer:
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :
Here, a = g
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
You might be interested in
The kinetic energy is .
Explanation:
The kinetic energy of an object is given by
where
K is the kinetic energy of the object
m is the mass of the object
v is the speed of the object
For the comet in this problem, we have:
is its mass
is the speed
First, we convert the speed from km/h to m/s:
Therefore, the kinetic energy of the comet is
Learn more about kinetic energy here:
#LearnwithBrainly
Answer:
vf₁ = 6.86 m/s , to the right
vf₂ = 2.96 m/s, to the right
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁= 0.220 kg : mass of object₁
m₂= 0.345 kg : mass of object₂
v₀₁ = 2.1 m/s ₁ , to the right : initial velocity of m₁
v₀₂= 6 m/s, to the right i :initial velocity of m₂
Problem development
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂
We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:
(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂
2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)
(2.1 - 6 ) = (vf₂ -vf₁)
-3.9 = (vf₂ -vf₁)
vf₂ = vf₁ - 3.9
vf₂ = vf₁ - 3.9 Equation (2)
We replace Equation (2) in the Equation (1)
2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)
2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)
2.532 + 1.3455 = (0.565)*vf₁
3.8775 = (0.565)*vf₁
vf₁ = (3.8775) / (0.565)
vf₁ = 6.86 m/s, to the right
We replace vf₁ = 6.86 m/s in the Equation (2)
vf₂ = 6.86 - 3.9
vf₂ = 2.96 m/s, to the right
Answer:
a) 6 times farther. b) 6 times longer.
Explanation:
Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:
vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s
In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:
vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s
Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:
If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:
On earth, this time could be calculated in the same way:
As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:
Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m
On earth, the distance travelled had been as follows:
Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m
⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00
b) As we have just found, the time of flight on the moon and on the earth are as follows:
tmoon = 11. 9 s
tearth = 1.98 s
⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0