Question

1. You are standing on the roof of a building, 46.0 m above the ground. your friend, who is 1.80 m tall, is standing next to the building. if your friend does not move, how long will it take for an egg to drop on his or her head?
A. 3.06
B. 3.12
C. 3.00
D. 9.00

2. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no predictable air drag. Compute the velocity of the sandbag at 0.250 s after it's release.
A. 2.55 m/s
B. 7.45 m/s
C. -2.45 m/s
D. 1.25 m/s

3. You are on a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball takes 2.40 s to reach the ground. How fast is the ball moving just as it lands? Neglect air resistance.
A. 1.12 m/s
B. 7.87 m/s
C. 3.55 m/s
D. 2.81 m/s

4. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. How far does it fall in 2.00 s?
A. 19.6 m
B. 49.6 m
C. 10.4 m
D. 99.2 m
5. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. What is it's speed after falling freely for 2.00 s?
A. 19.6 m/s
B. 34.6 m/s
C. 30.0 m/s
D. 4.6 m/s

6. Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter tree takes 2.0 s to reach the ground, how long will it take the other coconut to touch the ground?
A. 4.0 s
B. 2.8 s
C. 1.0 s
D. 1.4 s

7. One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given distance. A lander vehicle on a distant planet records the fact that it takes 3.17 s for a ball to fall freely 11.26 m, starting from rest. What is the acceleration due to gravity on that planet?
A. 3.55 m/s^2
B. 1.12 m/s^2
C. 2.24 m/s^2
D. 7.10 m/s^2

8. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no predictable air drag. Compute the position of the sandbag at 0.250 s after it's release.
A. 40.3 m
B. 39.7 m
C. 41.3 m
D. 40.9 m

9. Which statement regarding the idealized model of motion called free fall is true?
A. A freely falling object has a constant acceleration called the acceleration due to gravity.
B. The Idealized model of motion called free fall applies in cases where the distance of the fall is large compared with the radius of the astronomical body on which the fall occurs.
C. The effect of air resistance is factored into the equations of motion in the idealized model called free fall.
D. Free fall only models motion for objects that do not have an initial velocity in the upward direction.

10. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. What is the magnitude of it's velocity after falling 10.0 m?

Answer 1

1. C. 3.00

Equation: $S=\frac{1}{2}at^2$

where

S = 46.0 m - 1.80 m = 44.2 m is the space between the roof of the building and the head of your friend

$a=9.81 m/s^2$ is the acceleration of gravity

t is the time

Re-arranging the equation and substituting the numbers, we find the time:

$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.2 m)}{9.81 m/s^2}}=3.00 s$

2. A. 2.55 m/s

Equation: $v=u+at$

where

v is the final velocity of the sandbag

u=5.00 m/s is the initial velocity of the sandbag

a=-9.81 m/s^2 is the gravitational acceleration (with a negative sign since the direction is opposite to the initial velocity)

t=0.250 s is the time

Substituting into the equation, we find

$v=u+at=5.00 m/s+(-9.81 m/s^2)(0.25 s)=2.55 m/s$

3. B. 7.87 m/s

Equation: $v=u+at$

where

v is the final velocity of the ball

u=0 m/s is the initial velocity of the ball

a=3.28 m/s^2 is the gravitational acceleration

t=2.40 s is the time

Substituting into the equation, we find

$v=u+at=0 m/s+(3.28 m/s^2)(2.40 s)=7.87 m/s$

4. B. 49.6 m

Equation: $S=ut+\frac{1}{2}at^2$

where

S is how far the coin fell down

u=15 m/s is the initial velocity

t=2.00 s is the time

a=9.81 m/s^2 is the gravitational acceleration

Substituting into the equation,

$S=(15 m/s)(2.0 s)+\frac{1}{2}(9.81 m/s^2)(2.0 s)^2=49.6 m$

5. B. 34.6 m/s

Equation: $v=u+at$

where

v is the final velocity of the coin

u=15.0 m/s is the initial velocity of the coin

a=3.28 m/s^2 is the gravitational acceleration

t=2.00 s is the time

Substituting into the equation, we find

$v=u+at=15.0 m/s+(9.81 m/s^2)(2.00 s)=34.6 m/s$

6. B. 2.8 s

The height of the shorter tree is :

$S=\frac{1}{2}at^2=\frac{1}{2}(9.81 m/s^2)(2.0 s)^2=19.6 m$

The height of the longer tree is twice, so S=19.6 x 2=39.2 m

The time it takes for a coconut on the longer tree to fall down is

$t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(39.2 m)}{9.81 m/s^2}}=2.8 s$

7. C. 2.24 m/s^2

Equation: $S=\frac{1}{2}at^2$

where

S = 11.26 m is the distance covered by the ball

$a$ is the acceleration of gravity on the planet

t=3.17 s is the time

Re-arranging the equation and substituting the numbers, we find the acceleration:

$a=\frac{2S}{t^2}=\frac{2(11.26 m)}{(3.17 s)^2}=2.24 m/s^2$

8. D. 40.9 m

Position of the sandbag after time t: $S=S_0+ut+\frac{1}{2}at^2$

where

$S_0=40 m$ is the initial position

u=5.00 m/s is the initial velocity of the sandbag

a=-9.81 m/s^2 is the gravitational acceleration (with a negative sign since the direction is opposite to the initial velocity)

t=0.250 s is the time

Substituting into the equation, we find

$S=40 m+(5.0 m/s)(0.250 s)+\frac{1}{2}(-9.81 m/s^2)(0.250 s)^2=40.9 m$

9. The correct statement is:

A. A freely falling object has a constant acceleration called the acceleration due to gravity.

10. 20.5 m/s

Equation: $v^2 = u^2 +2aS$

where

v is the final velocity of the coin

u=15.0 m/s is the initial velocity

a=9.81 m/s^2 is the acceleration of gravity

S=10.0 m is the distance

Substituting numbers in the equation, we find

$v=\sqrt{u^2+2aS}=\sqrt{(15.0m/s)^2+2(9.81 m/s^2)(10.0 m)}=20.5 m/s$

Answer 2

1. C. 3.00

2. A. 2.55 m/s

3. B. 7.87 m/s

4. B. 49.6 m

5. B. 34.6 m/s

6. B. 2.8 s

7. C. 2.24 m/s^2

8. D. 40.9 m

9. A. A freely falling object has a constant acceleration called the acceleration due to gravity.

10. 20.5 m/s

Further explanation

1. You are standing on the roof of a building, 46.0 m above the ground. your friend, who is 1.80 m tall, is standing next to the building. if your friend does not move, how long will it take for an egg to drop on his or her head?

$t = \sqrt{\frac{2S}{a} } = \sqrt{\frac{2(46.0 m - 1.80 m)}{9.81 m/s^2} } = 3 s$

2. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no predictable air drag. Compute the velocity of the sandbag at 0.250 s after it's release.

$v = u + a*t \\v= 5.00 m/s+ (-9.81 m/s^2*0.250 s)\\v = 2.55 m/s$

3. You are on a planet where the acceleration due to gravity is known to be $3.28 m/s^2$. You drop a ball and record that the ball takes 2.40 s to reach the ground. How fast is the ball moving just as it lands? Neglect air resistance.

$v = u + a*t \\v= 0 m/s+ (3.28 m/s^2*2.40 s)\\v = 7.87 m/s$

4. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. How far does it fall in 2.00 s?

$S=ut+\frac{1}{2} at^2 = (15.0 m/s*2.00 s) + (\frac{1}{2} 9.81 m/s^2*(2.00 s)^2)\\S = 49.6 m$

5. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. What is it's speed after falling freely for 2.00 s?

$v = u +at = 15.0 m/s +(3.28 m/s^2 *2.00 s)\\v = 34.6 m/s$

6. Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter tree takes 2.0 s to reach the ground, how long will it take the other coconut to touch the ground?

$t = \sqrt{\frac{2S}{a} } = \sqrt{\frac{2*(19.6 *2)}{9.81 m/s^2} } = 2.8s$

7. One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given distance. A lander vehicle on a distant planet records the fact that it takes 3.17 s for a ball to fall freely 11.26 m, starting from rest. What is the acceleration due to gravity on that planet?

$a = \frac{2S}{t^2} = \frac{2*11.26 m}{(3.17 s)^2 } = 2.24 m/s^2$

8. A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no predictable air drag. Compute the position of the sandbag at 0.250 s after it's release.

$S = S_0 +ut+\frac{1}{2} at^2\\S = 40m + (5.00 m/s*0.250 s) + \frac{1}{2} (-9.81 m/s^2)*(0.250 s)^2\\S = 40.9 m$

9. Which statement regarding the idealized model of motion called free fall is true?

A. A freely falling object has a constant acceleration called the acceleration due to gravity.

10. A student throws a coin vertically downward from the top of a building. The coin leaves the thrower's hand with a speed of 15.0 m/s. What is the magnitude of it's velocity after falling 10.0 m?

$v = \sqrt{u^2+2aS} = \sqrt{(15.0 m/s)^2+2*9.81 m/s^2*10.0 m}\\v = 20.5 m/s$

Learn more

1. Learn more about the velocity brainly.com/question/10962624
2. Learn more about gravity brainly.com/question/1833803
3. Learn more       about  the acceleration brainly.com/question/1852158

Answer details

Grade:        9

Subject:  physics

Chapter: the acceleration

Keywords:  the velocity, constant speed, gravity, vertical,  falling freely