Answer:
a) The velocity of the ball when it hits the ground is -20.5 m/s.
b) To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.
Explanation:
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<em>A ball is thrown from a rooftop with an initial downward velocity of magnitude v0=2.9 m/s. The rooftop is a distance above the ground, h= 21 m. In this problem use a coordinate system in which upwards is positive.</em>
<em>(a) Find the vertical component of the velocity with which the ball hits the ground.</em>
<em>(b) If we wanted the ball's final speed to be exactly 27, 3 m/s from what height, h (in meters), would we need to throw it with the same initial velocity?</em>
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The equation of the height and velocity of the ball at any time "t" are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the ball at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = velocity of the ball at a time "t".
First, let´s find the time it takes the ball to reach the ground (the time at which h = 0)
h = h0 + v0 · t + 1/2 · g · t²
0 = 21 m - 2.9 m/s · t - 1/2 · 9.8 m/s² · t²
Solving the quadratic equation using the quadratic formula:
t = 1.8 s ( the other solution of the quadratic equation is rejected because it is negative).
Now, using the equation of velocity, let´s find the velocity of the ball at
t = 1.8 s:
v = v0 + g · t
v = -2.9 m/s - 9.8 m/s² · 1.8 s
v = -20.5 m/s
The velocity of the ball when it hits the ground is -20.5 m/s.
b) Now we have the final velocity and have to find the initial height. Using the equation of velocity we can obtain the time it takes the ball to acquire that velocity:
v = v0 + g · t
-27.3 m/s = -2.9 m/s - 9.8 m/s² · t
(-27.3 m/s + 2.9 m/s) / (-9.8 m/s²) = t
t = 2.5 s
The ball has to reach the ground in 2.5 s to acquire a velocity of 27.3 m/s.
Using the equation of height, we can obtain the initial height:
h = h0 + v0 · t + 1/2 · g · t²
0 = h0 -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²
-h0 = -2.9 m/s · 2.5 s - 1/2 · 9.8 m/s² · (2.5 s)²
h0 = 38 m
To acquire a final velocity of 27.3 m/s, the ball must be thrown from a height of 38 m.
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