The paintball player shoots horizontally at the speed of 75m/s from the height 1.5m above the ground. The ball misses its target and hits the ground
(a) For how many seconds is the ball in the air? (b) Find the maximum horizontal displacement
Answers
Answer:
Hey mate here is your answer...
Kepler's first law states that the orbit of a planet is an ellipse with a semimajor axis a and with the sun at one focus of the ellipse. A circular orbit has a semimajor axis equal to the radius of the circle, so a1=RE where RE is the radius of the earth's orbit; the eccentricity of a circle is 0. The other extreme is an ellipse with eccentricity 1 which is a straight line from the sun to the earth and so the semimajor axis for a "dropped earth" is a2=RE/2. Of course, this is not an orbit you will actually see in nature because there is no such thing as a point mass and the speed of the earth when the two point masses coincide would be infinite, but if we can cleverly deduce the period of this orbit, one-half that period will be the answer to your question.
Kepler's third law states that the square of the period T of an orbit is proportional to the cube of its semimajor axis, so T12/T22=a13/a23=RE3/(RE/2)3=8. So, T2=T1/√8=0.354 years. So, the time to go half a period is 0.177 years=64.6 days. I will let you put that into seconds.
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Explanation:
Given
Hight = 1 .5m
v = 75m/s
H = ut - gt^2
H = gt^2
t = sqrt (H/g)
t = sqrt75/10
t = sqrt7.5s
at horizontal
u = 75m/s
v = 0
v^2 = u^2 - 2gH
H = (u^2 - v^2 )/2g
H =75*75/20
H = 75*15/4
H = 1125/4
H = 281.25m