A small ball rolls horizontally off the edge of a tabletopthat is 1.20 m high. It strikes the floor at a point 1.52 m hori-zontally from the table edge. (a) How long is the ball in the air?(b) What is its speed at the instant it leaves the table?
Question
A small ball rolls horizontally off the edge of a tabletopthat is 1.20 m high. It strikes the floor at a point 1.52 m hori-zontally from the table edge. (a) How long is the ball in the air?(b) What is its speed at the instant it leaves the table?
Solution
(a) The time the ball is in the air can be calculated using the equation of motion for vertical displacement, which is given by:
h = 1/2 * g * t^2
where:
- h is the height of the table (1.20 m),
- g is the acceleration due to gravity (approximately 9.81 m/s^2), and
- t is the time the ball is in the air.
Rearranging the equation to solve for t gives:
t = sqrt((2*h)/g)
Substituting the given values gives:
t = sqrt((2*1.20 m)/(9.81 m/s^2)) = 0.49 s
So, the ball is in the air for approximately 0.49 seconds.
(b) The speed of the ball as it leaves the table can be calculated using the equation for horizontal displacement, which is given by:
d = v * t
where:
- d is the horizontal distance the ball travels (1.52 m),
- v is the speed of the ball as it leaves the table, and
- t is the time the ball is in the air (0.49 s, as calculated above).
Rearranging the equation to solve for v gives:
v = d/t
Substituting the given values gives:
v = 1.52 m / 0.49 s = 3.10 m/s
So, the speed of the ball as it leaves the table is approximately 3.10 m/s.
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