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vivado [14]
3 years ago
8

From the top of a tall building, you throw one ball straight up with speed v0 and one ball straight down with speed v0. (a) Whic

h ball has the greater speed when it reaches the ground? (b) Which ball gets to the ground first? (c) Which ball has a greater displacement when it reaches the ground? (d) Which ball has traveled the greater distance when it hits the ground?
Physics
1 answer:
Vladimir79 [104]3 years ago
7 0

Answer:

none

Explanation:

physics is fun

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A book falls off a shelf that is 10.0 m tall. What is the velocity at which the book hits the ground?
Elena L [17]

Answer:

14 m/s

Explanation:

The motion of the book is a free fall motion, so it is an uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. Therefore we can find the final velocity by using the equation:

v^2 = u^2 + 2gd

where

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

d = 10.0 m is the distance covered by the book

Substituting data, we find

v=\sqrt{0^2 + 2(9.8 m/s^2)(10.0 m)}=14 m/s

8 0
3 years ago
What is the speed of a car that travels 150 km in 3.00 hrs?
zhuklara [117]
You would do distance divided by speed. So 150÷3, which would equal 5km per hour.
4 0
3 years ago
Read 2 more answers
Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,
crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
2 years ago
if you dig a hole through the earth, from one hemisphere to the other, and drop a tennis ball into it, would it get stuck at the
tia_tia [17]

Answer:

Depending on which hemisphere it is, like western to eastern, It would most likely get stuck at the center. You would also have to put more things into thought like acceleration, velocity, and speed.

BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.

7 0
2 years ago
Read 2 more answers
Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an
VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given \Delta x=7\ m

\theta=45 \°

Also, \Delta y=(3.5-2)=1.5\ m

a_x=0\ and\ a_y=-9.81\ m/s^2

Let us say the velocity in the x-direction is v_x and in the y-direction is v_y. And acceleration in the x-direction is a_x and in the y-direction is a_y.

Also, \Delta x\ and\ \Delta y is distance covered in x and y direction respectively. And t is the time taken by the ball to hit the backboard.

We can write v_x=v_0cos(45)\ and\ v_y=v_0sin(45). Where v_0 is velocity of ball.

Now,

\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt

\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}

Also,

\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s.

Plugging this value in

t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}

t=1.06\ seconds

So, the time of flight of the ball is 1.06 seconds.

6 0
3 years ago
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