At the same moment when his teammate punts the ball, a football player starts fr
ID: 1274476 • Letter: A
Question
At the same moment when his teammate punts the ball, a football player starts from rest, at a point exactly alongside the kicker, and runs down the field with constant (non-zero) acceleration, toward the opponent who is waiting to catch the ball. If the ball leaves the kicker's foot at an angle of 76 degrees above the horizontal, what acceleration does the player need in order to arrive and tackle the opponent just as he catches the ball? Assume the ball is caught at the same height from which it was kicked. Ignore air resistance. Answer given is 2.44m/s.
Explanation / Answer
The time it takes for the ball to go a distance D is D/vx, where vx=v*cos(theta). Theta=76 degrees.
The time of flight of the ball is entirely dependent upon the y-direction kinematics. In the y-direction, to arrive back to where it started in height, delta-y=0.
delta-y=-0.5*g*t^2+vy*t
t=0 or t=2*vy/g (0 is obviously where it started)
The runner has only that much time to make it the distance D. His displacement is given as
delta-x_runner=D=0.5*a*t^2 (since he starts at x=0 with no velocity)
Since t=2*vy/g=D/vx, D=2*vy*vx/g=sin(2*theta)*v^2/g
D=0.5*a*t^2=sin(2*theta)*v^2/g
Now plug in the time again, 2*vy/g,
a=2*D/t^2=2*sin(2*theta)*v^2/g/(4*v^2*sin^2(theta)/g^2)
a=sin(2*theta)/(2*sin^2(theta))*g thus the units DO work
theta=76, g=9.81
a=2.445 m/s2 :)
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