What is systematic error
2 answers:
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Answer:
The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s
Explanation:
The given parameters of the ball are;
The initial speed of the ball = 15 m/s
The direction in which the ball is launched = 50° above the horizontal
The location of the other tennis player when the ball is launched = 10 m from the ball
The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched
The height at which the ball is hit back = 2.1 m above the height from which the ball is launched
The vertical position, 'y', at time, 't', of a projectile motion is given as follows;
y = (u·sinθ)·t - 1/2·g·t²
When y = 2.1 m, we have;
2.1 = (15·sin(50°))·t - 1/2·9.8·t²
∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0
Solving with the aid of a graphing calculator function, we get;
t = 0.199776187257 s or t = 2.14525782198 s
Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds
The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;
x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m
The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m
Therefore, we have;
The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s
The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m
The minimum average speed the other player has to move with, = d/t₂
∴ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s
The minimum average speed the opponent must move so that he is in position to hit the ball, ≈ 5.79 m/s.
Missing figure and missing details can be found here:
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
points in the opposite direction of the displacement of the cart, and where
therefore, the work done by the weight is
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Solution:
(a) The work done by the spring is given by
where k is the elastic constant of the spring and
(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
where the negative sign is given by the fact that
therefore, the work done by the weight is