A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 16.0 m/s,at an angle of θ = 51°, from the horizontal.
The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down (as shown in the Figure). Calculate how long the ball is in the air. ( g = 9.81 m/s2)
Vo = 16.0m/s[51o]
Yo = 16*sin51 = 12.43 m/s. = Vertical component of initial velocity.
Y = Yo + g*Tr = 0
Tr = -Yo/g = -12.43/-9.81 = 1.27 s. =
Rise time.
h = Yo*Tr + 0.5g*Tr^2 =
12.43*1.27 - 4.9*1.27^2 = 7.89 m. =
Max. ht.
h = ho - 0.5g*Tf^2 = 2 m
7.89 - 4.9*Tf^2 = 2
4.9Tf^2 = 7.89 - 2 = 5.89
Tf^2 = 1.20
Tf = 1.10 s. = Fall time.
Tr + Tf = 1.27 + 1.10 = 2.37 s. = Time
in air.
To calculate the time the ball is in the air, we can start by analyzing the vertical motion of the ball. We'll use the following kinematic equation for vertical motion:
Δy = v₀y * t + (1/2) * a * t²
Where:
Δy = vertical displacement (2.0 m)
v₀y = vertical component of the initial velocity (v₀ * sinθ)
t = time in the air
a = acceleration due to gravity (-9.81 m/s²)
First, let's find the vertical component of the initial velocity (v₀y). Using trigonometry:
v₀y = v₀ * sinθ
Substituting the given values:
v₀ = 16.0 m/s
θ = 51°
v₀y = 16.0 m/s * sin(51°)
Calculate v₀y:
v₀y ≈ 12.226 m/s
Now, let's use the kinematic equation and solve for time (t):
Δy = v₀y * t + (1/2) * a * t²
2.0 m = 12.226 m/s * t + (1/2) * (-9.81 m/s²) * t²
Rearrange the equation to solve for t:
(1/2) * (-9.81) * t² + 12.226 * t - 2.0 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Where:
a = (1/2) * (-9.81) = -4.905
b = 12.226
c = -2.0
Substitute the values into the quadratic formula:
t = (-12.226 ± √(12.226² - 4 * -4.905 * -2.0)) / (2 * -4.905)
Simplify the equation:
t = (-12.226 ± √(150 - 39.24)) / (-9.81)
t = (-12.226 ± √110.76) / (-9.81)
Now, solve for t using both the positive and negative root of the quadratic equation:
t₁ = (-12.226 + √110.76) / (-9.81)
t₂ = (-12.226 - √110.76) / (-9.81)
Calculate t₁:
t₁ ≈ 2.127 seconds
Calculate t₂:
t₂ ≈ -1.680 seconds
Since time cannot be negative, we discard the negative root. Therefore, the ball is in the air for approximately 2.127 seconds.
To find the time the ball is in the air, we can use the kinematic equation for vertical motion:
h = h0 + v0yt + 0.5gt^2
Where:
h = final height (2.0 m)
h0 = initial height (0 m)
v0y = vertical component of initial velocity (v0 * sin(θ))
g = acceleration due to gravity (-9.81 m/s^2)
t = time
Using the given values:
h = 2.0 m
h0 = 0 m
v0 = 16.0 m/s
θ = 51°
g = -9.81 m/s^2
To find the initial vertical velocity, v0y, we can use the equation:
v0y = v0 * sin(θ)
Plugging in the values:
v0y = 16.0 m/s * sin(51°)
v0y ≈ 12.23 m/s
Now, let's rearrange the first equation to solve for time:
t = (-v0y ± √(v0y^2 - 2g(h - h0))) / g
Plugging in the values:
t = (-12.23 m/s ± √((12.23 m/s)^2 - 2 * -9.81 m/s^2 * (2.0 m - 0 m))) / -9.81 m/s^2
Calculating this equation will give two possible values for time, since the ball can be caught on its way up or on its way down. However, we need to use the positive value of time, as the question states that the father catches the ball on its way down.
t = (-12.23 m/s + √((12.23 m/s)^2 - 2 * -9.81 m/s^2 * (2.0 m - 0 m))) / -9.81 m/s^2
Solve this equation to find the time the ball is in the air.