Asked by shmza
A place kicker must kick a football from a point 37.1 m from a goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked the ball leaves the ground with a speed of 19.8 m/s at an angle of 53° to the horizontal.
(a) By how much does the ball clear or fall short of clearing the crossbar?
(b) Does the ball approach the crossbar while still rising or while falling?
(a) By how much does the ball clear or fall short of clearing the crossbar?
(b) Does the ball approach the crossbar while still rising or while falling?
Answers
Answered by
Henry
Vo = 19.8m/s@53Deg.
Xo = hor = 19.8cos53 = 11.92m/s.
Yo = ver = 19.8sin53 = 15.81m/s.
a. t = (Yf - Yo) / g,
t(up) = (0 - 15.81) / -9.8 = 1.613s.
d=Vo*t + 4.9*t^2 = 12.753 - 3.05=9.703m
0 + 4.9t^2 = 9.703,
t^2 = 1.98,
t(dn) = 1.41s. to fall to 3.05m.
Dh = Xo(t(up)+t(dn)),
Dh = 11.92(1.41+1.613) = 36m=hor dist.
Falls short by: 37.1 - 36 = 1.1m.
b. While falling.
Xo = hor = 19.8cos53 = 11.92m/s.
Yo = ver = 19.8sin53 = 15.81m/s.
a. t = (Yf - Yo) / g,
t(up) = (0 - 15.81) / -9.8 = 1.613s.
d=Vo*t + 4.9*t^2 = 12.753 - 3.05=9.703m
0 + 4.9t^2 = 9.703,
t^2 = 1.98,
t(dn) = 1.41s. to fall to 3.05m.
Dh = Xo(t(up)+t(dn)),
Dh = 11.92(1.41+1.613) = 36m=hor dist.
Falls short by: 37.1 - 36 = 1.1m.
b. While falling.
Answered by
Anonymous
a. Falls short by 3 m
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