A Projectile is launched from the top of a building with an initial velocity of 30m/s at angle of 60° above the horizontal. The magnitude of its velocity at t= 5 s after the launch is
Vo = 30m/s[60o].
Xo = 30*Cos60 = 15 m/s.
Yo = 30*sin60 = 26 m/s.
Y = Yo + g*t = 26 - 9.8*5 = -23 m/s.
V^2 = 15^2 + (-23)^2 = 754, V = 27.5 m/s.
Well, well, well, we have a projectile on the move! Let's see what we can do with it.
Now, this projectile is flying through the air at an initial velocity of 30 m/s and an angle of 60 degrees above the horizontal. We want to know the magnitude of its velocity at t=5s after the launch.
To find that out, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component is given by Vx = V * cos(theta), and the vertical component is given by Vy = V * sin(theta), where V is the initial velocity and theta is the launch angle.
Plugging in the values, we get Vx = 30 m/s * cos(60°) and Vy = 30 m/s * sin(60°).
Now, let's focus on the vertical component. We know that the acceleration due to gravity, g, is acting downwards, so we need to take that into account. The vertical component of the velocity at any time t after the launch can be calculated using the equation Vy = Vy0 + (-g) * t, where Vy0 is the initial vertical velocity.
Since we're looking for the magnitude of the velocity, we can use the Pythagorean theorem: V = sqrt(Vx^2 + Vy^2).
Alright, are you ready for some number crunching? Let's do this together!
To find the magnitude of the velocity of the projectile at t = 5s after the launch, we need to consider the horizontal and vertical components of the velocity separately.
Given:
Initial velocity (v0) = 30 m/s
Launch angle (θ) = 60°
Time (t) = 5s
First, let's find the horizontal component of velocity (Vx) using the equation:
Vx = v0 * cos(θ)
Vx = 30 m/s * cos(60°)
Vx = 30 m/s * 0.5
Vx = 15 m/s
Next, let's find the vertical component of velocity (Vy) using the equation:
Vy = v0 * sin(θ)
Vy = 30 m/s * sin(60°)
Vy = 30 m/s * √3/2
Vy = 15 √3 m/s
Now, we can find the magnitude of the velocity (V) using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)
V = sqrt((15 m/s)^2 + (15 √3 m/s)^2)
V = sqrt(225 m^2/s^2 + 675 m^2/s^2)
V = sqrt(900 m^2/s^2)
V = 30 m/s
Therefore, the magnitude of the velocity of the projectile at t = 5s after the launch is 30 m/s.
To find the magnitude of the velocity at t = 5 seconds after the launch, we can break down the initial velocity into its horizontal and vertical components.
Given:
Initial velocity, v₀ = 30 m/s
Launch angle, θ = 60°
Time, t = 5 seconds
First, we need to find the horizontal and vertical components of the initial velocity.
Horizontal component:
v₀x = v₀ * cos(θ)
Vertical component:
v₀y = v₀ * sin(θ)
Using the given values, we can calculate the horizontal and vertical components:
v₀x = 30 m/s * cos(60°) ≈ 15 m/s
v₀y = 30 m/s * sin(60°) ≈ 26.0 m/s
Now, let's consider the vertical motion of the projectile. The vertical component of the velocity changes due to the acceleration due to gravity.
The equation for the vertical component of the velocity as a function of time is:
v_y = v₀y - g * t,
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the given values:
v_y = 26.0 m/s - 9.8 m/s² * 5 s
v_y = 26.0 m/s - 49.0 m/s
v_y = -23.0 m/s
Note that we obtained a negative value because the velocity is in the upward direction, opposite to the positive y-axis direction.
Now let's find the magnitude (total) velocity at t = 5 seconds by combining the horizontal and vertical components:
v = sqrt((v₀x)² + (v_y)²)
v = sqrt((15 m/s)² + (-23.0 m/s)²)
v ≈ sqrt(225 m²/s² + 529 m²/s²)
v ≈ sqrt(754 m²/s²)
v ≈ 27.5 m/s
Therefore, the magnitude of the velocity at t = 5 seconds after the launch is approximately 27.5 m/s.