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Answer:
The average speed of the fielder is 5.24 m/s
Explanation:
The position vector of the ball after it was hit can be calculated using the following equation:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem.
When the ball is caught, its position vector will be (see r1 in the figure):
r1 = (r1x, 0.873 m)
Then, using the equation of the position vector written above:
r1x = x0 + v0 · t · cos α
0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²
Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:
r1x = v0 · t · cos α
0.873 m = v0 · t · sin α + 1/2 · g · t²
Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:
0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²
0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²
Solving the quadratic equation:
t = 0.03 s and t = 5.72 s.
It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.
With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:
r1x = v0 · t · cos α
r1x = 37.2 m/s · 5.72 s · cos 49.3°
r1x = 1.39 × 10² m
The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.
The average velocity is calculated as the traveled distance over time, then:
average velocity = treveled distance / elapsed time
average velocity = 30.0 m / 5.72 s = 5.24 m/s
Answer:1.71 m/s
Explanation:
Given
mass of Susan
Inclination
Tension
coefficient of Friction
Resolving Forces Along x axis
where
since there is no movement in Y direction therefore
and
Thus
Work done by applied Force is equal to change to kinetic Energy
Answer:
The instantaneous speed of the object after the first five seconds is 12.5 m/s.
(C) is correct option.
Explanation:
Given that,
An object starts at rest. Its acceleration over 30 seconds.
We need to calculate the instantaneous speed of the object after the first five seconds
We know that,
Area under the acceleration -time graph gives speed.
According to figure,
Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.