1) A place-kicker must kick a football from a point 36.0 m (about 40 yards) from
ID: 1783607 • Letter: 1
Question
1) A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.8 m/s at an angle of 48.0° to the horizontal.
(a) By how much does the ball clear or fall short (vertically) of clearing the crossbar? (Enter a negative answer if it falls short.)
_______________m
2) A hiker walks 3.55 km south and then 2.25 km east, all in 2.80 hours.
(a) Calculate the magnitude (in km) and direction (in degrees south of east) of the hiker's displacement during the given time.
magnitude _____________ km
direction _______________° south of east
(b) Calculate the magnitude (in km/h) and direction (in degrees south of east) of the hiker's average velocity during the given time.
magnitude _____________ km/h
direction _______________° south of east
(c) What was her average speed (in km/h) during the same time interval?
_______________ km/h
Explanation / Answer
1)
along horizontal
________________
initial velocity v0x = v*cos(theta)
acceleration ax = 0
displacement x = R
from equation of motion
x = v0x*T+ 0.5*ax*T^2
x = v*costheta*T
T = x/(v*cos(theta))......(1)
along vertical
______________
v0y = v*sin(theta)
acceleration ay = -g = -9.8 m/s^2
from equation of motion
y-y0 = vy*T + 0.5*ay*T^2
y-y0 = (v*sin(theta)*x)/(v*cos(theta)) - (0.5*g*x^2)/(v^2*(cos(theta))^2)
y-y0 = x*tan(theta) - ((0.5*g*x^2)/(v^2*(cos(theta))^2))
X = 36 m
theta = 48
v = 23.8 m/s
y0 = 0
y = ?
y-y0 = 36*tan48 - (0.5*9.8*36^2/(23.8*cos48)^2)
y = 15 m
the ball will clear the cross bar by 15 - 3.05 = 11.95 m <<<----ANSWER
=======================
2)
(a)
displacement r1 = -3.55 j
displacement r2 = 2.25 i
total displacement r = r1 + r2 = 2.25i - 3.55 j
magnitude = sqrt(2.25^2+3.55^2) = 4.2 km
direction = tan^-1(-3.55/2.25) = 57.6 south of east
(b)
average velocity = total displacement /time
vavg = ( 2.25i - 3.55 j )/2.8 = 0.804 i - 1.27 j km/h
magnitude = sqrt(0.804^2+1.27^2) = 1.5 km/h
direction = tan^-1(1.27/0.804) = 57.6 south of east
(c)
total distance = 3.55 + 2.25 = 5.8 km
average speed = 5.8/2.8 = 2.1 km/h
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