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Eva8 [605]
3 years ago
9

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 fo

llows a parabolic path to the floor. How far along the floor is the landing spot from the table?
Physics
2 answers:
Mekhanik [1.2K]3 years ago
6 0

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.

sp2606 [1]3 years ago
5 0

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.

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