Final answer:
The horizontal distance between the edge of the table and the location where the ball lands can be calculated using the equation d = v*t. The time the ball is in the air is found using the equation d = ut + (1/2)at^2. The horizontal velocity when the ball leaves the table's edge can be calculated using the equation v = d / t. The speed of the ball when it hits the floor is found using the equation vf = vi + at.
Step-by-step explanation:
To find the horizontal distance between the edge of the table and the location where the ball lands, we can use the equation:
d = v*t
where d is the distance, v is the horizontal velocity, and t is the time.
First, let's calculate the time the ball is in the air.
Using the equation:
d = ut + (1/2)at^2
where d is the distance, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time, we can solve for t:
0.7 = 0 + (1/2)(-9.81)t^2
t^2 = (2*0.7) / 9.81
t = sqrt((2*0.7) / 9.81)
t = 0.37709s
Next, we'll calculate the horizontal velocity of the ball when it leaves the table's edge. Since there is no horizontal acceleration, the horizontal velocity remains constant. The horizontal distance traveled is the distance between the ramp and the edge of the table, which is 0.2m. Therefore, the horizontal velocity is:
v = d / t = 0.2 / 0.37709 = 0.5305m/s
Finally, to find the speed when the ball hits the floor, we can use the equation:
vf = vi + at
The initial vertical velocity is 0, the acceleration due to gravity is -9.81m/s^2, and the time t is 0.37709s. Solving for vf:
vf = 0 + (-9.81)(0.37709)
vf = -3.694m/s
The magnitude of the speed is the absolute value of the velocity, so the speed when the ball hits the floor is 3.694m/s.