A 0.23 kg ball of dough is thrown straight up into the air with an initial velocity of 12 m/s.Find the momentum of the ball of dough halfway to its maximum height on the way up.
KE at half height=
1/2 m v^2=1/2 (1/2 m 12^2)
or v= sqrt(1/2 144)=sqrt72
momentum at half height:
.23*sqrt72
check that.
what is sqrt??
Square Root
To find the momentum of the ball of dough halfway to its maximum height, we need to know its velocity at that point. The velocity of an object thrown vertically can be determined using the equation:
v = u - gt
Where:
v is the final velocity,
u is the initial velocity,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time.
Since we want to find the velocity halfway to the maximum height, we need to determine the time it takes for the ball of dough to reach that point.
The time it takes for an object to reach its maximum height can be calculated using the equation:
t = u / g
Where:
t is the time,
u is the initial velocity,
g is the acceleration due to gravity.
In this case, the initial velocity of the ball is 12 m/s, so:
t = 12 / 9.8
t ≈ 1.22 s
Therefore, it takes approximately 1.22 seconds for the ball of dough to reach its maximum height.
Now that we know the time it takes for the ball to reach halfway, we can substitute this value back into the equation for velocity to find its value at that point:
v = u - gt
v = 12 - (9.8 * 1.22)
v ≈ 0 m/s
The velocity at the halfway point is approximately 0 m/s.
To find the momentum, we use the equation:
momentum = mass * velocity
Given that the mass (m) of the ball of dough is 0.23 kg and the velocity (v) at the halfway point is 0 m/s, we can calculate the momentum:
momentum = 0.23 kg * 0 m/s
momentum = 0 kg*m/s
Therefore, the momentum of the ball of dough halfway to its maximum height on the way up is 0 kg*m/s.