A lovesick lad wants to throw a bag of candy and love notes into the open window of his girlfriend's bedroom 10 m above. Assuming it just reaches the window, he throws the love gifts at 60° to the ground:
a) At what velocity should he throw the bag?
b) How far from the house is he standing when he throws the bag?
eggs
a) Well, when it comes to impressing a lovesick lad, it's all about the velocity! To calculate the required velocity, we can break it down into two components: horizontal and vertical.
First, let's look at the vertical component. The bag needs to reach a height of 10 m, so we can use the formula: vf = vi + at. Since the final velocity (vf) when the bag reaches the window is 0 (since it stops), our equation becomes: 0 = vi + (-9.8)t. Since the initial velocity (vi) in the vertical direction is 0 (since he throws it upwards), it simplifies to: 0 = -9.8t. To find the time (t), we can use t = sqrt(2h/g), where h is the vertical distance (10 m) and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the numbers, we get t = sqrt((2*10)/9.8) ≈ sqrt(2.04) ≈ 1.43 seconds.
Now let's move on to the horizontal component. The horizontal distance traveled by the bag is what we need to figure out next. We know that the angle at which he throws the bag is 60°. Using trigonometry, we can find the horizontal distance (d) by using the equation: d = vi*cosθ*t, where vi is the initial velocity and θ is the launch angle. So, d = vi*cos(60)*t. Plugging in the values we have, d = vi*(1/2)*1.43.
Now, since he wants the bag to just reach the window, the horizontal distance should be equal to the distance from where he's standing to the house. Let's call the distance from the house "x". Therefore, x = vi*(1/2)*1.43.
b) So, to find how far from the house he is standing, we can rearrange the equation to solve for x. It becomes: x = (2 * d) / 1.43. Plugging in the known values, x = (2 * (vi*(1/2)*1.43)) / 1.43.
In summary:
a) To find the velocity, we can use the equation v = d/t, where v is the velocity, d is the horizontal distance, and t is the time. In this case, v = (vi*(1/2)*1.43) / 1.43.
b) To find the distance from the house, x = (2 * (vi*(1/2)*1.43)) / 1.43.
Now, let's hope he doesn't throw like a clown and hits the window instead of the neighbor's cat! 🤡
To solve this problem, we can break it down into two components: vertical and horizontal motion.
a) To find the velocity at which the lovesick lad should throw the bag, we can use the vertical motion component. We need to determine the initial vertical velocity (Vy) at which the bag should be thrown to reach a height of 10m.
We know that the vertical motion is affected by the acceleration due to gravity, which is approximately 9.8 m/s^2. The equation for the vertical motion can be represented as:
Δy = Vy*t + (1/2)*g*t^2
Where:
Δy = vertical displacement (10 m)
t = time of flight (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
Since the initial vertical velocity (Vy) is unknown, we can use a trigonometric relationship. The given angle of 60° is the angle of projection. In projectile motion, the vertical component of the initial velocity can be represented as:
Vy = V * sin(θ)
Where:
V = initial velocity
θ = angle of projection (60°)
Now we can use these equations to solve for the velocity (V). Substituting the given values into the equation:
10 m = V * sin(60°) * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying this equation yields a quadratic equation. Solving this equation will give us two values of time (t), one of which is the throwing time and the other is the fall time. We are interested in the throwing time.
Once we know the throwing time, we can substitute it back into the equation for the vertical component of the initial velocity (Vy) to find the velocity (V).
b) To find the distance the lad is standing from the house when he throws the bag, we need to find the horizontal component of the velocity (Vx). We can use the equation:
Vx = V * cos(θ)
Where:
V = initial velocity
θ = angle of projection (60°)
Once we find Vx, we can use the formula for horizontal motion:
Δx = Vx * t
Where:
Δx = horizontal distance (unknown)
t = time of flight (throwing time obtained from part a)
By substituting the values into the equation, we can find the distance from the house where the lad should stand to throw the bag.
Using these equations, we can find the answers to parts a) and b) of the question.
I dont get it :(((
the equation of motion for an object thrown from (0,0) at an angle θ with velocity v is
y(x) = -g/(2v2 cos2θ) x2 + xtanθ
the range (where y=0 again) is
r = v2 sin2θ/g
the maximum height reached is
h = v2 sin2θ/2g
So, we know that
h = 10
θ = 60°
10 = v2 (3/4)/(2*9.8)
10 = .038 v2
v2 = 263.16
v = 16.22
The range is twice the distance to the balcony, so the balcony is at half the range:
r = 16.222 sin(120)/9.8
= 263.09 * √3/2 / 9.8
= 23.24
so, he stood 11.62m from the house