Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:
Answer:
527 ° C
Explanation:
For black body radiation, the power emitted per unit area is directly proportional to the absolute temperature raised to the fourth power:
j = k T⁴
First, convert 127° C to Kelvins:
127° C + 273.15 = 400.15 K
Now find the constant of variation:
2×10⁵ = k (400.15)⁴
k = 7.80×10⁻⁶
Finally, solve for the temperature at the new rate of radiation:
32×10⁵ = (7.80×10⁻⁶) T⁴
T = 800.3
The temperature is 800 K, or about 527 ° C.
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Before we find impulse, we need to find the initial and final momentum of the ball.
To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.
Values we know:
acceleration(a) - 9.81m/s^2 [down]
initial velocity(vi) - 0m/s
distance(d) - 1.25m [down]
This equation can be used to find final velocity:
Vf^2 = Vi^2 + 2ad
Vf^2 = (0)^2 + (2)(-9.81)(-1.25)
Vf^2 = 24.525
Vf = 4.95m/s [down]
Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.
What we know:
a = 9.81m/s^2 [down]
d = 0.600m [up]
vf = 0m/s
Vf^2 = Vi^2 + 2ad
0^2 = Vi^2 + 2(-9.81)(0.6)
0 = Vi^2 + -11.772
Vi^2 = 11.772
Vi = 3.43m/s [up]
Now to find impulse given to the ball by the floor we find the change in momentum.
Impulse = Momentum final - momentum initial
Impulse = (0.120)(3.43) - (0.120)(-4.95)
Impulse = 1.01kgm/s [up]
