Lily throws a ball at an angle of 53.13 degrees above the horizontal from a building that is 10m high. The ball has a speed of 12m/s, .6 seconds after she throws it. What was the initial speed of the ball? This is a vector problem and uses variables such as vx, vy, vyi, delta y and delta x
Vinitial = Vi
vx = x velocity = constant = Vi cos 53.13
vyi = Vi sin 53.13
at t = .6
vx =vx still
vy = vyi - g t
so
V at.6=12 = sqrt {vx^2+ (Vyi-9.81*.6)^2]
To find the initial speed of the ball, we need to break down the initial velocity into horizontal and vertical components. Let's assume the initial speed is denoted as "v" and the angle of projection is denoted as "θ".
Given:
- The angle of projection, θ = 53.13 degrees
- The initial height, h = 10m
- The time after release, t = 0.6s
- The initial vertical velocity, vy = ?
First, we can find the vertical component of the initial velocity using the equation:
vy = v * sin(θ)
To find the horizontal component of the initial velocity, we can use the equation:
vx = v * cos(θ)
However, since the projectile is thrown horizontally from a building, the horizontal component of velocity remains constant throughout the motion. Therefore, we can ignore it for now.
Next, we need to find the initial vertical velocity, vyi. We know the vy (vertical component of velocity) 0.6 seconds after release, and we can use that to calculate vyi using the equation:
vyi = vy + g * t
where g is the acceleration due to gravity (approximately equal to 9.8 m/s^2).
Now, we can determine vyi:
vyi = vy + g * t
= v * sin(θ) + g * t
From the problem statement, we are given vy when t = 0.6s as vy = 0.
0 = v * sin(θ) + g * t
Solving for v:
v = -g * t / sin(θ)
Substituting the known values:
v = -9.8 m/s^2 * 0.6 s / sin(53.13°)
Now we can calculate v using a scientific calculator:
v ≈ -9.8 m/s^2 * 0.6 s / sin(53.13°) ≈ -5.94 m/s
Since the initial speed cannot be negative, we can take the absolute value of v to obtain the initial speed of the ball:
Initial speed = |v| ≈ |-5.94 m/s| ≈ 5.94 m/s
Therefore, the initial speed of the ball is approximately 5.94 m/s.