1. A bullet fired at an angle θ = 60o with a velocity of 20 m/s. Acceleration due to gravity is 10 m/s2. What is the time interval to reach the maximum height?
Known :
The initial velocity of bullet (vo) = 20 m/s
Angle (θ) = 60oC
Acceleration due to gravity (g) = 10 m s–2
Wanted : The time interval to reach the maximum height
Solution :
The initial velocity at the horizontal direction (x axis) :
vox = vo cos 60o = (20)(0.5) = 10 m/s
The initial velocity at the vertical direction (y axis) :
voy = vo sin 60o = (20)(0.5√3) = 10√3 m/s
The time interval to reach the maximum height, calculated using this equation :
vty = voy + g t
vty = the final velocity in the vertical direction = the final velocity at the highest point = 0 m/s
voy = the initial velocity at the horizontal direction = 10√3 m/s
g = acceleration due to gravity = 10 m/s2
t = time interval
The time interval :
vty = voy + g t
0 = 10√3 – 10 t
10√3 = 10 t
t = 10√3 / 10
t = √3 seconds
[irp]
2. An object projected at an angle. The height of the object is the same when the time interval = 1 second and 3 seconds. What is the time interval the object in air.
Solution :
The object in the air for 4 seconds.
3. An aircraft is moving horizontally with a speed of 50 m/s. At the height of 2 km, an object is dropped from the aircraft. Acceleration due to gravity = 10 m/s2, what is the time interval before the object hits the ground.
Known :
Height = 2 km = 2000 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted : The time interval (t)
Solution :
h = 1/2 g t2
2000 = 1/2 (10) t2
2000 = 5 t2
t2 = 2000/5 = 400
t = √400 = 20 seconds
[irp]
4. A kicked football leaves the ground at an angle θ = 45o with the horizontal has an initial speed of 25 m/s. Determine the distance of X. Acceleration due to gravity is 10 m/s2.
Known :
Initial speed (vo) = 25 m/s
Acceleration due to gravity (g) = 10 m/s2
Angle (θ) = 45o
Wanted : X
Solution :
The horizontal component of the initial velocity :
vox = vo cos θ = (25 m/s)(cos 45o) = (25 m/s)(0.5√2) = 12.5√2 m/s
The vertical component of the initial velocity :
voy = vo sin θ = (25 m/s)(sin 45o) = (25 m/s)(0.5√2) = 12.5√2 m/s
Projectile motion could be understood by analyzing the horizontal and vertical component of the motion separately. The x motion occurs at constant velocity and the y motion occurs at constant acceleration of gravity.
Time in the air (t) :
The time in air calculated with the equation of the upward vertical motion.
Choose upward direction as positive and downward direction as negative.
Known :
The initial velocity (vo) = 12.5√2 m/s (upward direction, positive)
Acceleration due to gravity (g) = -10 m/s2 (downward direction, negative)
Height (h) = 0
Wanted : Time interval (t)
Solution :
h = vo t + 1/2 g t2
0 = (12.5√2) t + 1/2 (-10) t2
0 = 12.5√2 t – 5 t2
12.5√2 t = 5 t2
12.5√2 = 5 t
t = 12.5√2 / 5
t = 2.5√2 seconds
[irp]
The horizontal distance (X) :
Calculated using the equation of the uniform linear motion with constant velocity.
Known :
Velocity (v) = 12.5√2 m/s
Time interval (t) = 2.5√2 seconds
Wanted : Distance
Solution :
d = v t = (12.5√2)(2.5√2) = (12.5)(2.5)(2) = 62.5 meters
5. An object projected upward at an angle θ = 30o with the horizontal has an initial speed of 20 m/s. Acceleration due to gravity is 10 m/s2. Determine the maximum height.
Known :
The initial velocity (vo) = 20 m/s
Acceleration due to gravity (g) = 10 m/s2
Angle (θ) = 30o
Wanted : The maximum height
Solution :
First, find the vertical component of the initial velocity (voy) :
voy = vo sin 30o = (20)(sin 30o) = (20)(0.5) = 10 m/s
Calculate the maximum height. Choose upward direction as positive and downward direction as negative.
Known :
Acceleration due to gravity (g) = -10 m/s2 (downward direction, negative)
The vertical component of the initial velocity (voy) = 10 m/s (upward direction, positive)
Velocity at the maximum height (vty) = 0
Wanted : The maximum height (h)
Solution :
vt2 = vo2 + 2 g h
02 = 102 + 2 (-10) h
0 = 100 – 20 h
100 = 20 h
h = 100/20
h = 5 meters
The maximum height is 5 meters.
6. An object is thrown at a certain elevation angle. The height of the object same after 1 second and 3 seconds. Determine time in air.
A. 3.6 s
B. 4.0 s
C. 5.6 s
D. 6.4 s
Solution
Time in air = 4 seconds.
The correct answer is B.
7. An aircraft is moving horizontally with the speed of 50 m/s. When the aircraft at the height of 2 km, an object free fall from the aircraft. Determine the type of the motion.
A. Free fall motion
B. Floating motion
C. Horizontal motion
D. Projectile motion
Solution :
The object is dropped from the moving plane because it has the same speed as the plane’s speed, that is 50 m/s. Movement of objects is not like free fall motion but parabolic motion. The case is the same as you are dropping objects from inside a moving car.
The correct answer is D.
