What is wrong about the picture below
1 answer:
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Answer:
2.2 s
Explanation:
Hi!
Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by
We are looking for a time t for which x(t) = 0
Solving for t:
Rounding at the first decimal:
t = 2.2 s
Answer:
Your friend has to wait 0.26 s after you throw the ball to start running.
Explanation:
The equation that gives the position vector of the ball is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal positon
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity
The equation of displacement of your friend is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of your friend at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.
Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:
y = y0 + v0 · t ·sin α + 1/2 · g · t²
-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m
Solving the quadratic equation:
t = 1.71 s
Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):
x= x0 + v0 · t · cos α
x = 0m + 9.00 m/s · 1.71 s · cos 33°
x = 12.9 m
Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.
Let´s see, how much time it takes your friend to run that distance:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)
x = 1/2 · a · t²
1.9 m = 1/2 · 1.80 m/s² · t²
Solving for t
t = 1.45 s
Then, since the time of flight of the ball is 1.71 s, your friend has to wait
1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.
