Final answer:
To demonstrate the speed and direction of a particle after 3 seconds, the initial horizontal and vertical velocities are calculated, and these components are used to find the final velocity using the Pythagorean theorem, confirming that it is 18√5 m/s at an angle of tan⁻¹(1/2) to the horizontal.
Step-by-step explanation:
A student wishes to calculate the velocity of a particle after 3 seconds that was initially projected from ground level at a speed of 60 m/s and an angle of tan⁻¹(4/3) to the horizontal. We begin by finding the initial horizontal and vertical components of the velocity. The angle of projection can be converted from the inverse tangent form to degrees or radians, but we can directly use the trigonometric ratios for calculations.
The initial horizontal velocity (Ux) is given by:
Ux = initial speed × cos(angle) = 60 m/s × (3/5) = 36 m/s
The initial vertical velocity (Uy) is given by:
Uy = initial speed × sin(angle) = 60 m/s × (4/5) = 48 m/s
Since there is no horizontal acceleration, the horizontal velocity remains constant at 36 m/s. The vertical velocity, however, is affected by gravity. After 3 seconds, the vertical velocity (Vy) is:
Vy = Uy - (gravity × time) = 48 m/s - (9.8 m/s² × 3s) = 19.6 m/s
We can now calculate the resultant velocity (V) using the Pythagorean theorem:
V = √(Ux² + Vy²) = √(36² + 19.6²) m/s = 18√5 m/s
To find the angle above the horizontal (θ), we take the inverse tangent of the ratio of vertical to horizontal velocities:
tan(θ) = Vy / Ux = (19.6 / 36)
θ = tan⁻¹(1/2)
Therefore, the velocity of the particle after 3 seconds is indeed 18√5 m/s at an angle of tan⁻¹(1/2) above the horizontal.