Physics: A small steel ball bearing with a mass of 15.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.21 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 31.0 deg from the horizontal.

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A small steel ball bearing with a mass of 15.0

am2016832

18.01.2022

Horizontal Distance d=2.14 m

Explanation:

Given data

Mass m=15.0 g =0.015 kg

Height h=1.21 m

Angle α=31.0°

To find

Horizontal distance d

Solution

The horizontal distance as given as:

$d_{distance}=((vi_{initail-velocity})^{2} /g_{gravity} )*Sin2\alpha$

First We need to find the initial velocity Vi

$v_{i}=\sqrt{2gh}\\v_{i}=\sqrt{2(9.8m/s)(1.21m)}\\v_{i}=4.87m/s$

Substitute the given values and initial velocity to find horizontal distance:

$d=(vi^{2} /g)*Sin2\alpha\\ d=(\frac{(4.87m/s)^{2} }{9.8m/s^{2} } )Sin(2*31.0)\\d=2.14m$

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