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:0.58 m

Explanation:

The initial velocity of the ball is u = 2.0 m/s

The height of the table is, h = 1.0 m

The ball falls in vertical direction under acceleration due to gravity.

Time taken for ball to hit the floor:

h= ut + 0.5gt² ( from the equation of motion)

1.0 m=2.0 m/s × t+0.5 × 9.8 m/s²× t²

Solving this for t,

t = 0.29 s ( we have neglected the negative value of t)

In the same time, the ball would cover a horizontal distance of :

s = u t

⇒s = 2.0 m/s×0.29 s = 0.58 m

Thus, the landing spot is 0.58 m away from the table.

Answer 2

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.