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Salsk061 [2.6K]
3 years ago
6

Aiden takes 0.10s to throw a baseball, which leaves his hand with a velocity of 49m/s. The balls acceleration is

Physics
1 answer:
Brums [2.3K]3 years ago
7 0
Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
a =  \frac{v}{t}
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
a =  \frac{49}{0.1}  \\ a =  \frac{490}{1}  \\ a = 490
The acceleration is 490 m/s^2
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A ball is droped from a height of 16m how much time will pass before the ball hits the ground​
sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

g = 9.8m{s}^{-2}

By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

square root both sides, we obtain

\implies t= \sqrt{3.265}

t=1.8s

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

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