Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 24 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
What is the maximum height the ball goes above the ground?
Vi = 24 sin 32 = 12.72 m/s
v = Vi + a t
so at top
0 = 12.72 - 9.81 t
t = 1.3 seconds rising
h = 1.5 + 12.72(1.3) - .5(9.81)(1.3^2)
h = 1.5 + 16.5 - 8.3
h = 9.7 m
To find the maximum height the ball reaches, we need to analyze the projectile motion. Since the only force acting on the ball is gravity (assuming no air resistance), the motion can be divided into horizontal and vertical components.
The horizontal component is unaffected by gravity and remains constant throughout the motion. So, we can find the time it takes for the ball to reach Sarah's position by using the horizontal distance traveled.
The vertical motion is affected by gravity and can be analyzed using the kinematic equations of motion. The key equation for finding the maximum height is:
y = y0 + v0y * t - (1/2) * g * t^2
where:
y = height above the ground
y0 = initial height above the ground
v0y = initial vertical component of velocity
t = time
g = acceleration due to gravity (typically -9.8 m/s^2)
First, let's find the time it takes for the ball to reach Sarah's position by using the horizontal distance:
Horizontal distance = v0 * cos(angle) * time
In this case, the horizontal distance is zero since the ball is caught at the same horizontal position where it was thrown. Therefore, we can conclude that the time it takes for the ball to reach Sarah is zero.
Now, let's find the maximum height:
0 = y0 + v0y * t - (1/2) * g * t^2
Since the time is zero, the term containing t^2 will be zero:
0 = y0 + v0y * t
Now, we can solve for y0, which represents the maximum height:
y0 = -(v0y * t)
Since the time is zero, y0 will also be zero. This means that the maximum height above the ground is zero.
Therefore, the ball does not go above the ground in this case.
To find the maximum height the ball reaches, we can use the kinematic equation for vertical motion.
The equation is:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
- y is the height above the ground
- y0 is the initial height above the ground (1.5 m)
- v0y is the initial vertical velocity (in this case, the vertical component of the initial velocity)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time
First, we need to find the initial vertical velocity, v0y. This can be calculated using the initial velocity and the angle of projection.
v0y = v0 * sin(θ)
Where:
- v0 is the initial speed of the ball (24 m/s)
- θ is the angle of projection (32 degrees)
v0y = 24 * sin(32°)
Using a calculator, v0y ≈ 12.95 m/s
Next, we need to find the time it takes for the ball to reach its maximum height. At the highest point of the ball's trajectory, the vertical component of the velocity is zero.
v = v0y - g * t
0 = 12.95 - 9.8 * t
Solving for t:
9.8 * t = 12.95
t = 12.95 / 9.8
Using a calculator, t ≈ 1.32 s
Now that we have the time, we can find the maximum height reached by substituting the values into the equation:
y = y0 + v0y * t - (1/2) * g * t^2
y = 1.5 + (12.95) * (1.32) - (0.5) * (9.8) * (1.32)^2
Using a calculator, y ≈ 8.4 m
Therefore, the maximum height the ball reaches above the ground is approximately 8.4 meters.