1. C. 3.00
Equation: ![S=\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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
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](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2S%7D%7Ba%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2844.2%20m%29%7D%7B9.81%20m%2Fs%5E2%7D%7D%3D3.00%20s)
2. A. 2.55 m/s
Equation: ![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
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](https://tex.z-dn.net/?f=v%3Du%2Bat%3D5.00%20m%2Fs%2B%28-9.81%20m%2Fs%5E2%29%280.25%20s%29%3D2.55%20m%2Fs)
3. B. 7.87 m/s
Equation: ![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
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](https://tex.z-dn.net/?f=v%3Du%2Bat%3D0%20m%2Fs%2B%283.28%20m%2Fs%5E2%29%282.40%20s%29%3D7.87%20m%2Fs)
4. B. 49.6 m
Equation: ![S=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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](https://tex.z-dn.net/?f=S%3D%2815%20m%2Fs%29%282.0%20s%29%2B%5Cfrac%7B1%7D%7B2%7D%289.81%20m%2Fs%5E2%29%282.0%20s%29%5E2%3D49.6%20m)
5. B. 34.6 m/s
Equation: ![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
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](https://tex.z-dn.net/?f=v%3Du%2Bat%3D15.0%20m%2Fs%2B%289.81%20m%2Fs%5E2%29%282.00%20s%29%3D34.6%20m%2Fs)
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](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2%3D%5Cfrac%7B1%7D%7B2%7D%289.81%20m%2Fs%5E2%29%282.0%20s%29%5E2%3D19.6%20m)
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](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2S%7D%7Ba%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2839.2%20m%29%7D%7B9.81%20m%2Fs%5E2%7D%7D%3D2.8%20s)
7. C. 2.24 m/s^2
Equation: ![S=\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
S = 11.26 m is the distance covered by the ball
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](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2S%7D%7Bt%5E2%7D%3D%5Cfrac%7B2%2811.26%20m%29%7D%7B%283.17%20s%29%5E2%7D%3D2.24%20m%2Fs%5E2)
8. D. 40.9 m
Position of the sandbag after time t: ![S=S_0+ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3DS_0%2But%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
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](https://tex.z-dn.net/?f=S%3D40%20m%2B%285.0%20m%2Fs%29%280.250%20s%29%2B%5Cfrac%7B1%7D%7B2%7D%28-9.81%20m%2Fs%5E2%29%280.250%20s%29%5E2%3D40.9%20m)
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](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B2aS)
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](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2aS%7D%3D%5Csqrt%7B%2815.0m%2Fs%29%5E2%2B2%289.81%20m%2Fs%5E2%29%2810.0%20m%29%7D%3D20.5%20m%2Fs)