Question

5) A ball rolls off the edge of a horizontal table top 4m high. It strikes the floor at a point 5m horizontally away from the edge of the table. How long was the ball in the air in seconds? A/2/√3 B/3/√5 C/1/√5 D/2/√5​

Answer 1

To calculate the time it takes for the ball to fall from the table to the floor, we can use the formula: time = (2 * distance) / acceleration due to gravityIn this case, the distance is the height of the table, which is 4m, and the acceleration due to gravity is approximately 9.8m/s^2. So the calculation would be:time = (2 * 4) / 9.8 = 0.816 secondsTherefore, the ball was in the air for about 0.816 seconds.

Answer 2

To solve this problem, we can use the equations of motion for an object in free fall. We'll assume that the ball was launched horizontally from the edge of the table with an initial velocity of zero.

Let's denote the time the ball was in the air as 't' seconds.

We know that the vertical displacement of the ball is equal to the height of the table, which is 4 meters. Using the equation for vertical displacement:

s = ut + (1/2)at^2

where:

s = vertical displacement (4 meters)

u = initial vertical velocity (0 m/s, as the ball is launched horizontally)

a = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)

t = time in seconds

Plugging in the values, we have:

4 = 0t + (1/2)(-9.8)t^2

4 = (-4.9)t^2

Dividing both sides by -4.9, we get:

t^2 = -4/4.9

Since time cannot be negative, we discard the negative sign:

t^2 = 4/4.9

Taking the square root of both sides:

t = sqrt(4/4.9) = 2/√4.9

So, the time the ball was in the air is approximately 2/√4.9 seconds.

Comparing the answer choices given:

A) A/2/√3

B) 3/√5

C) 1/√5

D) 2/√5

Among the choices, D) 2/√5 is the closest approximation to 2/√4.9. Therefore, the answer is D) 2/√5.