A ball rolls horizontally off the edge of a tabletop that is 1.60 m high. It strikes the floor at a point 1.58 m horizontally away from the table edge. (Neglect air resistance.)How long was the ball in the air? What was its speed at the instant it left the table?
The time in the air depends on height.
h= 1/2 g t^2 solve for t.
speed= horizontal distance/t
To find the time the ball was in the air, we can use the equation of motion for vertical motion:
h = (1/2)gt^2
where h is the height (1.60 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearrange the equation to solve for t:
t^2 = (2h)/g
t^2 = (2 * 1.60 m) / 9.8 m/s^2
t^2 = 0.3265 s^2
To find t, take the square root of both sides:
t = sqrt(0.3265 s^2)
t ≈ 0.571 s
Therefore, the ball was in the air for approximately 0.571 seconds.
To find the speed of the ball at the instant it left the table, we can use the horizontal distance and time:
speed = horizontal distance / t
speed = 1.58 m / 0.571 s
speed ≈ 2.76 m/s
Therefore, the speed of the ball at the instant it left the table was approximately 2.76 m/s.