6) Jessica and Eric are playing catch. Jessica tosses a ball to Eric at an initial velocity of 40 feet per second at an angle of 45° from a height of 4 feet. Jessica is 30 feet from Eric. How high above the ground will the ball be when it gets to Eric? Round all answers to the nearest hundredths.
First, we need to find how long it will take for the ball to travel from Jessica to Eric. We can use the formula:
time = distance / velocity
The distance between Jessica and Eric is 30 feet, and the velocity of the ball is 40 feet per second. Therefore:
time = 30 / 40
time = 0.75 seconds
Next, we can use the horizontal and vertical components of the initial velocity to calculate the vertical displacement of the ball during the 0.75 seconds it takes to travel from Jessica to Eric. The horizontal component of the initial velocity is:
vx = v * cos(theta)
vx = 40 * cos(45°)
vx = 28.28 feet per second
The vertical component of the initial velocity is:
vy = v * sin(theta)
vy = 40 * sin(45°)
vy = 28.28 feet per second
We can use the following kinematic equations for motion with constant acceleration:
vf = vi + at
d = vi*t + 0.5*a*t^2
vf^2 = vi^2 + 2ad
Since the ball is only affected by gravity in the vertical direction, we can use the following values for the vertical motion:
vi = 28.28 feet per second (upward)
vf = ?
a = -32.2 feet per second squared (downward acceleration due to gravity)
t = 0.75 seconds
d = ?
Using the first equation:
vf = vi + at
vf = 28.28 - 32.2(0.75)
vf = 3.02 feet per second (downward)
Using the second equation:
d = vi*t + 0.5*a*t^2
d = 28.28(0.75) + 0.5(-32.2)(0.75)^2
d = 12.61 feet (upward)
Therefore, the ball will be 12.61 feet above the ground when it gets to Eric.
To solve this problem, we can break it down into horizontal and vertical components. Let's start by finding the initial horizontal and vertical velocities of the ball:
Given:
Initial velocity (v) = 40 ft/s
Initial launch angle (θ) = 45°
Horizontal velocity (vx) can be calculated using the formula:
vx = v * cos(θ)
vx = 40 * cos(45°)
vx ≈ 40 * 0.7071
vx ≈ 28.284 ft/s
Vertical velocity (vy) can be calculated using the formula:
vy = v * sin(θ)
vy = 40 * sin(45°)
vy ≈ 40 * 0.7071
vy ≈ 28.284 ft/s
Next, we'll calculate the time it takes for the ball to reach Eric. Since Eric is 30 feet away, we only need to consider the horizontal distance:
Given:
Horizontal distance (d) = 30 ft
Horizontal velocity (vx) = 28.284 ft/s
Time (t) can be calculated using the formula:
t = d / vx
t = 30 / 28.284
t ≈ 1.0607 s
Now, let's find the vertical displacement (height) when the ball reaches Eric. We can use the following kinematic equation:
Height (h) = vy * t + 0.5 * g * t^2
Given:
Initial vertical position (y0) = 4 ft
Vertical velocity (vy) = 28.284 ft/s
Time (t) = 1.0607 s
Acceleration due to gravity (g) = -32.174 ft/s^2 (we consider downward direction as negative)
Height (h) = 28.284 * 1.0607 + 0.5 * (-32.174) * (1.0607^2)
Height (h) ≈ 30.13 ft
Therefore, the ball will be approximately 30.13 feet above the ground when it reaches Eric.
To find how high above the ground the ball will be when it gets to Eric, we can use the equations of projectile motion.
The initial vertical velocity (Vy) of the ball can be found by multiplying the initial velocity (v) by the sine of the launch angle (θ):
Vy = v * sin(θ)
In this case, v = 40 feet per second and θ = 45°. Plugging in these values, we get:
Vy = 40 * sin(45°)
Using the sine of 45° (which is √2/2), we can calculate:
Vy ≈ 40 * √2/2
Vy ≈ 20√2 feet per second
Next, we need to find the time it takes for the ball to reach Eric. We can use the horizontal distance (range) and the horizontal component of the velocity (Vx) to find the time (t):
range = Vx * t
In this case, the range is given as 30 feet, and the horizontal velocity (Vx) can be found by multiplying the initial velocity (v) by the cosine of the launch angle (θ):
Vx = v * cos(θ)
Plugging in the values, we have:
30 = Vx * t
Vx = 40 * cos(45°)
Using the cosine of 45° (which is √2/2), we can calculate:
Vx ≈ 40 * √2/2
Vx ≈ 20√2 feet per second
Substituting the value of Vx back into the equation, we have:
30 = 20√2 * t
To find t, we can rearrange the equation:
t = 30 / (20√2)
t ≈ 1.06 seconds
Now that we know the time it takes for the ball to reach Eric, we can find how high above the ground the ball will be.
We can use the equation of vertical displacement (h) without time:
h = Vy * t + 0.5 * g * t^2
where g is the acceleration due to gravity (32.2 ft/s^2).
Plugging in the values:
h = (20√2) * 1.06 + 0.5 * 32.2 * (1.06)^2
Simplifying, we get:
h ≈ 21.21 + 18.245
h ≈ 39.45 feet
Therefore, the ball will be approximately 39.45 feet above the ground when it reaches Eric.