On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25m/s at an angle of 35.0 above the horizontal. If she is in flight for 0.616sec, how high above the water was she when she let go of the rope?
Well, let's swing into action and calculate her height!
Firstly, we know that her initial velocity can be divided into its horizontal and vertical components. The horizontal component won't affect her height, so we'll focus on the vertical component.
Her initial vertical velocity (Vy) can be calculated using trigonometry: Vy = V * sin(theta), where V is the initial velocity and theta is the launch angle. Plugging in the values, we get:
Vy = 2.25 m/s * sin(35.0)
Now, the time of flight (t) is given as 0.616 seconds. Since the girl doesn't change her vertical velocity during her flight, we can use this to determine the vertical displacement (height, h).
The equation for vertical displacement under constant vertical velocity is h = Vy * t. Plugging in the values we already have, we get:
h = 2.25 m/s * sin(35.0) * 0.616 s
Now, we can calculate the height!
However, I must warn you, this calculation might make your brain feel like it's swinging in circles. So take a deep breath and brace yourself. Ready?
The calculated height (h) is approximately 0.88 meters. So, the young girl was about 0.88 meters above the water when she let go of the rope.
Now, imagine if all physics problems were this exciting! Maybe we should have a physics circus instead? The Flying Mathletes, perhaps?
To solve this question, we need to use the equations of motion for projectile motion. Let's break down the information given in the question:
Initial velocity (v₀) = 2.25 m/s at an angle of 35.0° above the horizontal.
Time in flight (t) = 0.616 seconds.
We need to find the height (h) at which the girl let go of the rope.
First, we need to analyze the vertical motion separately from the horizontal motion. We can calculate the initial vertical velocity (v₀y) using the initial velocity (v₀) and the angle (θ) as follows:
v₀y = v₀ * sin(θ)
= 2.25 m/s * sin(35.0°)
= 1.30 m/s
Next, we can use the equation for vertical displacement to find the height above the water (h):
h = v₀yt - (1/2)gt²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Let's plug in the values and calculate:
h = (1.30 m/s * 0.616 s) - (1/2)(9.8 m/s²)(0.616 s)²
= 0.79984 m - 1.89212 m
= -1.09228 m
Since the height cannot be negative, we can conclude that the girl was approximately 1.09 meters above the water when she let go of the rope.
Her vertical velocity component at release was 2.25*sin 35 = 1.29 m/s
Solve the following equation for altitude vs. time to get the initial height H:
y = H + 1.29 t - (g/2) t^2
When t = 0.616 s, y = 0
Use that to solve for H