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a hiker throws a ball at an angle of 21 degrees above the horizontal from a hill 21.0m high?
the hiker's height is 1.750 m
the magnitudes of the horizontal and vertical components are 14.004 m/s and 5.376 m/s respectively
find the distance between the base of the hill and the point where the all hits the ground
please show your work
consider the hiker's height while calculating the answer
If you ignore air resistance and just work with gravity, the time to hit the ground will be found by:
y=y0+u*t+1/2*a*t^2
where,
u=5.376 m/s
a=g=-9.807 m/s^2
y=0 m
y0=1.750+21.0=22.75 m (or 22.8 if you respect significant figures)
Solve the quadratic: t=(-b+-sqrt(b^2-4*a*c))/(2*a)
t=(-5.376 - sqrt(5.376^2 - 4*1/2*(-9.807)*22.75) / (2*1/2*(-9.807)) = 2.7708
This is the time the ball will hit the ground, so find how far it moves in x by this time (moves in x at constant speed).
x = vx*t = 14.004*2.7708 = 38.81 m
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