A ball is thrown from the ground onto a roof of height 15 m from a distance of 7 meters away as shown in the diagram. The maximum height of the ball's trajectory is 3.4 meters above the top of the roof.
Find the initial horizontal component of the velocity.
??????
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(a) Find the required initial vertical component of the velocity, Vy.
Vy = m/s *
19 OK
HELP: The ball is at its maximum height. What does this tell us about Vf?
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(b) Find the time for the ball to reach maximum height.
t = s *
1.936 OK
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(c) Find the time for the ball to fall from the maximum height to the rooftop.
t = s *
.83 OK
2007-10-11 11:37:50 · 1 answers · asked by SEXXYSuzzy 1 in Science & Mathematics ➔ Physics
See the ref. With g = 9.81,
v0y = sqrt(2g*(ymax-y0) = 19.0002105251495
t1 = sqrt(2(ymax-y0)/a) = 1.93682064476549
t2 = sqrt(2(ymax-yt)/a) = .83256845631734
t = t1+t2 = 2.76938910108283
v0x = range/t = 2.52763325935782
v0 = sqrt(v0x^2+v0y^2) = 19.167601046918
theta = arctan(v0y/v0x) = 82.4223289948109 deg
2007-10-14 03:57:28 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋