A placekicker must kick a football from a point 36.0 m (about 40 yards) from the

Question

A placekicker 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 20.2 m/s at an angle of 52.0 to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (Enter a negative answer if it falls short.) 5.01833 It may be helpful to first determine the time required for the ball to reach the goal. m (b) Does the ball approach the crossbar while still rising or while falling? rising O falling Need Help? Read It l Master It Submit Answer Save Progress

Explanation / Answer

(a)

PROJECTILE

along horizontal

initial velocity vox = vo*costheta

ax = 0

from equation of motion

x-Xo = vox*T+ 0.5*ax*T^2

x-X0 = vo*costheta*T

T = (x-X0)/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -g

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

y-y0 = (vo*sintheta*(x-x0))/(vo*costhetra) + (0.5*ay*(x-x0)^2)/(vo^2*(costheta)^2)

y-y0 = (x-X0)*tantheta + ((0.5*ay*(x-X0)^2)/(vo^2*(costheta)^2))

y0 = 0 m

y = 3.05 m

x-xo = 36

theta = 52

v0 = 20.2 m/s

y = (36*tan52) - (0.5*9.8*36^2/(20.2^2*(cos52)^2))

y = 5.02 m

by the time ball reaches the bar it is at a height of 5.02 m

part(a)

5.02-3.05 = 1.97 m   <<<<<-------answer

part(b)

time taken t = x/vox = 36/(20.2*cos52) = 2.89 s

vy = voy + ay*t

vy = 20.2*sin52 - (9.8*2.89) = -12.4 m/s

part(b)

falling <<<---------answer

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