Solutions to Practice Questions from Module 6:
1.a)If an object falls from a resting height of 490 m, how long does it remain in the air?
Known:vi = 0 m/s
a = g = -9.81 m/s2
y = -490 m
Unknown:
t = ? / Solution:
The object will remain in the air for 10. seconds.
b)If the object above is thrown directly upwards with a velocity of 8.2 m/s, how long does it remain in the air?
Known:vi = +8.2 m/s
a = g = -9.81 m/s2
y = -490 m/s
Unknown:
t = ? / Solution:
Use the quadratic equation to solve for t:
The object will remain in the air for 12 seconds.
2.A pitched baseball is thrown horizontally a distance of 18.3 m at 44.8 m/s
a)How long does the ball take to reach home plate?
Known:vix = vfx = 44.8 m/s
x = 18.3 m
viy = 0
a = g = 9.81 m/s2
Unknown:
t = ? / Solution:
It takes the ball 0.409 seconds to reach home plate.
b)How far does the ball drop during its flight?
Known:vix = vfx = 44.8 m/s
x = 18.3 m
viy = 0
a = g = 9.81 m/s2
t = 0.4085 s
Unknown:
y = ? / Solution:
The ball dropped downwards 0.819 meters during its flight.
3.A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 0.32 m below the height from which it was thrown. How far away is the player from the board?
Known:vix = vfx = 12.4 m/s
viy = 0
a = g = 9.81 m/s2
y = -0.32 m
Unknown:
x = ? / Solution:
Use vertical components to solve for time:
Now time to solve for horizontal distance:
The player is 10. meters away from the dart board.
4.A stone is thrown horizontally from a cliff 78.4 m height. If it lands 32 m from the base of the cliff, what was the release velocity?
Known:x = 32 m
viy = 0 m/s
y = -78.4 m/s
a = g = 9.81 m/s2
Unknown:
vx = ? / Solution:
Use the vertical components to solve for time:
Now solve for the horizontal velocity:
The stone was thrown with a horizontal velocity of 8.0 m/s
5.A mortar shell is fired at an angle of 53° with a speed of 98 m/s.
a)How long is the shell in the air?
b)How far does it travel before hitting the ground?
c)How high does it reach?
Known:vi = 98 m/s @ 53°
y = 0 m
a = g = 9.81 m/s2
Unknown:
t = ? / Solution:
Determine horizontal and vertical initial speeds:
The shell was in the air for 16 seconds.
Unknown:
x = ? / Solution:
The shell travels 940 meters before hitting the ground.
Unknown:
ymax = ? / Solution:
Because the parabolic trajectory is symmetrical, the shell will reach its maximum height at the halfway point. This will occur at the halfway time of 8 seconds.
The shell reaches a maximum height of 310 meters.
6.A cannon is fired at 30.0° above the horizontal with a velocity of 210. m/s from the edge of a cliff 125 m high. Calculate where the cannonball lands on the level plain below.
Known:vi = 210 m/s @ 30°
y = -125 m
a = g = 9.81 m/s2
Unknown:
x = ? / Solution:
Determine horizontal and vertical initial speeds:
Determine the time it took for the cannonball to land:
Use the quadratic equation to solve for t:
Use t = 22.5375 seconds to solve for x:
The cannonball landed 4100 meters away from the base of the cliff.
7.An athlete throws 7.3 kg shot-put with an initial speed of 14 m/s at a 42° angle to the horizontal. Calculate the distance traveled. The shot leaves the shot-putter’s hand at a height of 2.2 m above the ground.
Known:m = 7.3 kg (extraneous info.)
vi = 14 m/s @ 42°
y= -2.2 m
a = g = -9.81 m/s2
Unknown:
x = ? / Solution:
Determine horizontal and vertical initial speeds:
Determine the time it took for the cannonball to land:
Use the quadratic equation to solve for t:
Use t = 2.1213 seconds to solve for x:
The shot-put traveled a distance of 22 meters.