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A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

bottom of the ramp is at a height of h2 = 1.69 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.17 m.

1 answer:

Yuliya22 [3.3K]1 month ago

8 0

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

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

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

Next, we can find the ball's velocity

Using the kinetic energy formula

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

By applying the conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

We substitute the values into the equation

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

$d =vt$

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

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