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Elixer
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Homework Statement
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult.Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00o. How much farther on the opposite floor would it have landed if the downward angle were , instead, 8.00o?
(Source - Fundamentals of PHYSICS, Halliday, Resnick,Walker, 8th edition, chapter 4, QS.33)
Homework Equations
This is a case of projectile motion , thus,
say, initial velocity = u,
displacement in vertical direction = h
displacement in horizontal direction = s
time of flight = t
h = uyt + 0.5ayt2
s = uxt + 0.5axt2
The Attempt at a Solution
CASE 1:angle = 18o
uy = usin(18o) = 6.18m/s
-2.3 = 6.18t -4.9t2
t = 1.562s
ux= ucos(18o) = 19.02 m/s
so, s1 = 19.02 * 1.562 = 29.71 m
CASE 2:angle = 8o
uy = usin(8o) = 2.78m/s
-2.3 = 2.78t-4.9t2
t = 1.025s
ux = ucos(8o) = 19.81m/s
so, s2= 19.81*1.025 = 20.31m
Well, I am confused.
Thank you.