Asked by Michaela
A ball is projected horizontally from the edge of a table that is 1.15 m high, and it strikes the floor at a point 1.70 m from the base of the table.
(a) What is the initial speed of the ball?
m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?
m
(a) What is the initial speed of the ball?
m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?
m
Answers
Answered by
Henry
h = Vo*t + 0.5g*t^2 = 1.15m,
0 + 4.9t^2 = 1.15,
t^2 = 0.235,
t = 0.484s = Time in flight.
a. Dh = Xo * t = 1.70m, = Hor distance.
0.484Xo = 1.70,
Xo = 3.5m/s. = Initial hor velocity.
b. tan45 = Y/Xo,
Y = Xo*tan45 = 3.5 * 1 = 3.5m/s = ver.
comp. of velocity.
d = (Vf^2 - Vo^2) / 2g.
d = ((3.5)^2 - 0) / 19.8 = 0.625m.
h = 1.15 - 0.625 = 0.525m. above the floor.
0 + 4.9t^2 = 1.15,
t^2 = 0.235,
t = 0.484s = Time in flight.
a. Dh = Xo * t = 1.70m, = Hor distance.
0.484Xo = 1.70,
Xo = 3.5m/s. = Initial hor velocity.
b. tan45 = Y/Xo,
Y = Xo*tan45 = 3.5 * 1 = 3.5m/s = ver.
comp. of velocity.
d = (Vf^2 - Vo^2) / 2g.
d = ((3.5)^2 - 0) / 19.8 = 0.625m.
h = 1.15 - 0.625 = 0.525m. above the floor.
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