Solved Examples on Projectile Motion For Vertical Velocity
Example 1: A football player throws a ball with an initial velocity of 25 m/s in a direction, making an angle of 45° horizontal. Determine the vertical velocity of the ball if the time interval is 10 seconds. (g = 9.8 m/s2)
Solution:
Given data:
Initial velocity (u) = 25 m/s
Angle (θ) = 45°
Time interval (t) = 10 seconds
Acceleration due to gravity (g) = 9.8 m/s2
We have,
The vertical velocity in the projectile motion of a particle is given by:
Vy = Uy – gt
Vy = U sinθ – gt
= 25 × sin 45° – 9.8 × 10
= 17.68 – 98
= –80.32 m/s
Hence, the vertical velocity of the ball is –80.32 m/s.
Example 2: Determine the vertical velocity of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 30 m/s at an angle of 30°, and it hit the ground after 8 seconds. (g = 10 m/s2)
Solution:
Given data,
Initial velocity (u) = 30 m/s
Time interval = 8 sec
Angle (θ) = 30°
Acceleration due to gravity = 10 m/s2
We have,
The vertical velocity in the projectile motion of a particle is given by:
Vy = uy – gt = u sinθ – gt
= 30 × sin 30° – 10 × 8
= 30 × ½ – 80 {sin 30° = ½}
= 15 – 80 = –65 m/s
Hence, the vertical velocity of the ball is –65 m/s.
Example3: Determine the vertical velocity of the ball that is launched from some distance, making the projectile motion. The ball was thrown with an initial velocity of 40 m/s at an angle of 55°, and it hit the ground after 13 seconds. (g = 9.8 m/s2)
Solution:
Given data,
Initial velocity (u) = 40 m/s
Time interval = 13 sec
Angle of projection (θ) = 55°
Acceleration due to gravity = 9.8m/s2
We have,
The vertical velocity in the projectile motion of a particle is given by:
Vy = uy – gt = u sinθ – gt
= 40 × sin 55° – 9.8 ×13
= 40 × (0.8192) – 127.4 {sin 55° = 0.8192}
= 32.768 – 127.4 = –94.632 m/s
Hence, the vertical velocity of the ball is ––94.632 m/s.
Example 4: What is the initial velocity at which an object is thrown from some distance, making the projectile motion? The angle of projection is 45°, the vertical velocity is -32 m/s, and the object hit the ground after 8 seconds. (g = 10 m/s2)
Solution:
Given data,
Vertical velocity (u) = -32 m/s
Time interval = 8 sec
The angle of projection (θ) = 45°
Acceleration due to gravity = 10m/s2
We have,
The vertical velocity in the projectile motion of a particle is given by:
Vy = uy – gt = u sinθ – gt
-32 = u × sin 45° – 10 ×8
-32 = u × (1/√2) – 80 {sin 45° = 1/√2}
u/√2 = 80 – 32 = 48
u = 48√2 = 67.88 m/s
Hence, the initial velocity of the object is 67.88 m/s.
Example 5: A ball is launched at an initial velocity of 28 m/s and the vertical velocity of the ball and the time interval are –75 m/s and 9 seconds. Now, determine the angle that the ball forms with the horizontal. (g = 10 m/s2)
Solution:
Given data,
Vertical velocity (Vy) = –75 m/s
Time interval = 9 sec
Initial velocity (u) = 28 m/s
Acceleration due to gravity = 10 m/s2
We have,
The vertical velocity in the projectile motion of a particle is given by:
Vy = uy – gt = u sinθ – gt
⇒ –75 = 28 × sin θ – 10 × (9)
⇒ –75 = 28sin θ – 90
⇒ 28 sin θ = –75 + 90 = 15
⇒ sin θ = 15/28 = 0.7589
⇒ θ = sin-1 (15/28) = 32.39°
Hence, the angle of projection is 32.39°.
Projectile Motion For Vertical Velocity
The projectile motion of an object is the motion of an object that is launched into space. The object that is launched into space is known as a projectile, and the path traveled by it is called the trajectory of a projectile. Some real-life examples of a projectile are a javelin thrown by an athlete, a bullet fired from a gun, an arrow released from a bow, the launching of missiles, etc. When an object is launched into space, the only force that acts on it is gravity. There are also other forces like air resistance that act on a projectile, but their impact on it is minimal compared to gravity. Some initial force must be imparted to an object for it to become a projectile and remember that the object tossed straight upward is not considered a projectile. When an object is launched into the air at an angle with respect to the horizontal and with an initial velocity, it travels in a curved path under the influence of acceleration due to gravity. The path of a projectile is a predictable path that has a parabolic shape.
The above figure represents the projectile motion of an object that was launched at the point “O” (point of projection) with an initial velocity of “u” and at an angle θ. The object travels in a parabolic path and reaches a maximum height of “H” while being decelerated by gravity. As the object travels up, its vertical component of velocity decreases and becomes zero as it reaches maximum height, and then it will drop until it touches the ground due to gravitational pull. The distance between the point of projection (O) and the landing point (B) is called the range (R), and the time taken by the object to reach from O to B is called the time of flight.
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