Respuesta :
Given Information:
Launch angle of projectile = 30°
Initial velocity = V₀ = 30 m/s
Acceleration due to gravity = g = 10 m/s²
Required Information:
Angle with the horizontal after 1.5 sec = ?
Answer:
The angle of the projectile to the horizontal after t = 1.5 seconds is 0°
Step-by-step explanation:
The horizontal component of the velocity is given by
[tex]Vx = V_0 \cos(\theta)[/tex]
Where V₀ is the initial velocity and θ is the launch angle
The vertical component of the velocity is given by
[tex]Vy = V_0 \sin(\theta) - gt[/tex]
Where V₀ is the initial velocity, θ is the launch angle, g is the acceleration due to gravity and t is the time.
So after t = 1.5 sec
The horizontal component of the velocity is
[tex]Vx = V_0 \cos(\theta) \\\\Vx = 30 \cos(\30) \\\\Vx = 30 \times 0.866\\\\Vx = 25.981 \: m/s[/tex]
And the vertical component of the velocity is
[tex]Vy = V_0 \sin(\theta) - gt \\\\Vy = 30 \sin(30) - 10 \times 1.5 \\\\Vy = 30(0.5) - 10 \times 1.5 \\\\Vy = 15 - 15 \\\\Vy = 0 \: m/s \\\\[/tex]
The angel is
[tex]\tan(\theta) = \frac{0}{25.981} \\\\\theta= \tan^{-1}( \frac{0}{25.981}) \\\\\theta= 0[/tex]
Therefore, the angle of the projectile to the horizontal after t = 1.5 seconds is 0°
