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.

Answer 1

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.