The initial velocity of the ball is 55.125 m/s.
<h3>Initial velocity of the ball</h3>
The initial velocity of the ball is calculated as follows;
During upward motion
h = vi - ¹/₂gt²
h = vi - 0.5(9.8)(3²)
h = vi - 44.1 ----------------- (1)
During downward motion
h = vi + ¹/₂gt²
h = 0 + 0.5(9.8)(1.5)²
h = 11.025 ----------- (2)
solve (1) and (2) together, to determine the initial velocity of the ball
11.025 = vi - 44.1
vi = 11.025 + 44.1
vi = 55.125 m/s
Thus, the initial velocity of the ball is 55.125 m/s.
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Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
when the ground is very hot and the air is cool.
Explanation:
The hot earth warms a layer of air right above the ground. Light is refracted as it passes through the cool air and onto the hot air sheet (bent). A coating of very warm air near the earth bends the light from the sky almost into a U-shaped bend.
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