A projectile is fired with an initial velocity of 24 m/s at 40 degrees above the horizontal.What is the velocity of the projectile at its highest point?
At the highest point, the vertical velocity component is zero. All that remains is the horizontal velocity component, which remains constant at
24 cos 40 = 18.39 m/s
during flight.
To find the velocity of the projectile at its highest point, we need to separate the initial velocity into its horizontal and vertical components.
The horizontal component can be found using the formula:
Vx = V * cos(θ)
Where:
Vx: Horizontal component of velocity
V: Initial velocity
θ: Launch angle
In this case, the initial velocity (V) is given as 24 m/s, and the launch angle (θ) is given as 40 degrees. Plugging in these values:
Vx = 24 * cos(40)
Using a calculator, we find:
Vx ≈ 18.36 m/s (rounded to two decimal places)
Now, to find the vertical component of velocity, we can use the formula:
Vy = V * sin(θ)
Using the same values:
Vy = 24 * sin(40)
Again, using a calculator we find:
Vy ≈ 15.42 m/s (rounded to two decimal places)
At the highest point of the projectile's trajectory, the vertical component of velocity becomes zero. However, the horizontal component remains constant throughout the entire motion.
Therefore, the velocity of the projectile at its highest point can be determined by taking only the horizontal component of the initial velocity, which is approximately 18.36 m/s.