Question

A ball rolls with a speed of 2.0m/s across a level table that is 1.0m above the floor. Upon reaching the edge of the table it follows a parabolic path to the floor. How far along the floor is the landing spot from the table?

Answer 1

Answer:

The distance covered by the ball is 0.9 m.

Explanation:

Given that,

Speed of ball = 2.0 m/s

Distance = 1.0 m

We need to calculate the time

Using equation of motion

$h = ut+\dfrac{1}{2}gt^2$

Here, u = initial velocity=0

$t=\sqrt{\dfrac{3h}{g}}$

Where, h = distance

g = acceleration due to gravity

t = time

Put the value into the formula

$t=\sqrt{\dfrac{2\times1.0}{9.8}}$

$t=0.45\ sec$

We need to calculate the distance of the ball

Using formula of velocity

$d=v\times t$

$d=2.0\times0.45$

$d=0.9\ m$

Hence, The distance covered by the ball is 0.9 m.

Answer 2

Using the equation of motion and the speed - distance relation, the distance traveled by the ball is 0.9 meters.

Using the motion equation :

• S = ut + 0.5gt²

• S = distance ; g = 9.8 m/s ; t = time

Initial Velocity, u = Zero (0)

Substituting the values into the equation :

1 = 0(2) + 0.5(9.8)t²

1 = 0 + 4.9t²

4.9t² = 1

t² = 1/4.9

t² = 0.204

t = √0.204

t = 0.4517 seconds

Distance traveled = speed × time

Distance traveled = 2 m/s × 0.4517

Distance traveled = 0.90 m

Therefore, the distance in which the ball lands is 0.90 meters

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