Question 1191828: A player kicks a football at an angle of 35° with the horizontal and with an initial velocity of 14.8 m/s. A second player standing at a distance of 26.0 m from the first player in the direction of the kick starts running to meet the ball at the instant it is kicked.
(a) At what rate should the second player accelerate to catch the ball?
(b) What would be the speed of the second player upon catching the ball?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
t = elapsed number of seconds
x = horizontal distance the ball is from the kicker
y = vertical distance the ball is from the kicker
distances are in meters
vx = horizontal speed
vy = initial vertical speed
Speeds are in m/s.
Draw a right triangle. Mark the angle of elevation to be 35 degrees.
The hypotenuse of this triangle is the initial velocity 14.8 m/s.
The horizontal leg and vertical leg are vx and vy respectively.
Using trig ratios, we can say:
cos(theta) = adjacent/hypotenuse
cos(35) = (horizontal speed)/(diagonal speed)
cos(35) = (vx)/(14.8)
vx = 14.8*cos(35)
vx = 12.123450
and also,
sin(theta) = opposite/hypotenuse
sin(35) = (vertical speed)/(diagonal speed)
sin(35) = (vy)/(14.8)
vy = 14.8*sin(35)
vy = 8.488931
We can use those speeds to get
x = vx*t = 12.123450*t
y = vy*t - 0.5*g*t^2 = 8.488931*t - 0.5*g*t^2
where g = 9.81 m/s^2 approximately is the acceleration of gravity
For more info, check out the projectile motion equations.
Let's plug in that approximate value of g
y = 8.488931*t - 0.5*g*t^2
y = 8.488931*t - 0.5*9.81*t^2
y = 8.488931*t - 4.905*t^2
y = -4.905*t^2 + 8.488931*t
Now let's find out when the ball will hit the ground.
This occurs when y = 0.
y = -4.905*t^2 + 8.488931*t
-4.905*t^2 + 8.488931*t = y
-4.905*t^2 + 8.488931*t = 0
t(-4.905t + 8.488931) = 0
t = 0 or -4.905t + 8.488931 = 0
t = 0 or -4.905t = -8.488931
t = 0 or t = -8.488931/(-4.905)
t = 0 or t = 1.730669
We already know the ball is on the ground at the initial time t = 0 seconds (when it was kicked), so we'll ignore that solution.
The only solution is roughly t = 1.730669 seconds.
This is the amount of time the runner has to catch the ball.
Plug this t value into the x position equation found earlier.
x = 12.123450*t
x = 12.123450*1.730669
x = 20.981679
This is the horizontal distance the ball travels if we ignore wind resistance.
Accounting for wind resistance will greatly complicate the problem.
The runner is 26 meters ahead of the kicker and needs to get to the 20.981679 marker. The runner needs to turn back toward the kicker to run 26-20.981679 = 5.018321 meters.
To catch the ball in time, they need to run about 5.018321 meters in roughly 1.730669 seconds.
I'm assuming the runner does not get a head start, which means the runner's initial velocity is vi = 0
So,
a = acceleration
a = (x - vi*t)/(0.5*t^2) ... variation of a kinematics equation
a = (5.018321 - 0*1.730669)/(0.5*(1.730669)^2)
a = 3.350892
Answer: Approximately 3.350892 m/s^2
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Part (b)
Use the values mentioned in the previous part.
velocity = acceleration*time ... formula works only if vi = 0
velocity = 3.350892*t
velocity = 3.350892*1.730669
velocity = 5.799285
Answer: 5.799285 m/s approximately
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