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insens350 [35]
3 years ago
9

A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

Physics
1 answer:
noname [10]3 years ago
4 0

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration (g=9.8 m/s^2, acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

s=ut+\frac{1}{2}gt^2

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

g=9.8 m/s^2

t is the time of flight

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
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Answer:

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Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

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Look at the attached graphic

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F₁x= 33.4 *sin23.8° = 13.48 N

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Magnitude of the equilibrant (Fe)

F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2}  }

F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2}  }

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