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 2.25 m/s at an angle of 35 degrees above the horizontal. If she is in flight for 0.616 s, how high above the water was she when she let go of the rope?
Your name changes on each post. Are you answer grazing? Appears so. Never trust a person who changes their name on each question.
= 5.65m.
There are a whole mess of these questions all basically the same this morning. Read any old answer.
u = speed * cos angle
Vi = speed * sin angle
v = Vi - 9.81 t
v = 0 at top
h = Hi + Vi t - 4.9 t^2
https://www.mathsisfun.com/quadratic-equation-solver.html
range = u t
To determine the height above the water when the girl let go of the rope, we can use the equations of projectile motion. Here's how you can find the answer step by step:
Step 1: Break down the initial velocity into its horizontal (Vx) and vertical (Vy) components. The horizontal component (Vx) remains constant throughout the motion, while the vertical component (Vy) changes due to gravity. The initial velocity can be separated as follows:
Vx = Vi * cos(angle)
Vy = Vi * sin(angle)
Given that the initial velocity (Vi) is 2.25 m/s and the angle (angle) is 35 degrees, we can substitute these values into the equations to find the horizontal and vertical components.
Vx = 2.25 m/s * cos(35°)
Vy = 2.25 m/s * sin(35°)
Step 2: Determine the time of flight. The time of flight (t) is given as 0.616 s.
t = 0.616 s
Step 3: Calculate the height above the water. Since the motion is in a vertical direction, we only need to focus on the vertical component and use the equation:
h = Vy * t + (1/2) * g * t^2
In this equation, g represents the acceleration due to gravity (approximately 9.8 m/s²).
By substituting the values we know:
h = (2.25 m/s * sin(35°)) * 0.616 s + (1/2) * (9.8 m/s²) * (0.616 s)^2
Simplifying this equation will give you the height (h) above the water when she let go of the rope.