A ball is thrown at 22m/s at 45 degrees to the horizontal. a fence 5 feet tall is located 100 meters away. Does the ball make it over the fence?
constant horizontal speed = 22 cos 45
so time to fence = t = 100 /(22 cos 45)
so how high is it at time t?
h = initial height assume 0 + [ 22 sin 45] t - (1/2)(9.81) t^2
That gives you h in meters. Convert to feet to see if it is higher than five feet.
To determine if the ball makes it over the fence, we need to find the maximum height the ball reaches and compare it to the height of the fence.
Let's break down the motion of the ball into two components: horizontal and vertical.
1. Horizontal component:
The initial velocity in the horizontal direction (Vx) is given by:
Vx = velocity * cos(angle)
Vx = 22 * cos(45 degrees)
Vx = 22 * √2 / 2
Vx = 22 * 0.7071
Vx ≈ 15.56 m/s
The horizontal distance traveled (dx) can be calculated using the equation:
dx = Vx * time
Given that the distance to the fence is 100 meters, we can rearrange the equation to solve for time:
time = dx / Vx
time = 100 / 15.56
time ≈ 6.42 seconds
2. Vertical component:
The initial velocity in the vertical direction (Vy) is given by:
Vy = velocity * sin(angle)
Vy = 22 * sin(45 degrees)
Vy = 22 * √2 / 2
Vy = 22 * 0.7071
Vy ≈ 15.56 m/s
Using the formula for vertical displacement, we can calculate the maximum height (h) reached by the ball:
h = (Vy^2) / (2 * gravity)
h = (15.56^2) / (2 * 9.8)
h ≈ 12.79 meters
Comparing the maximum height of the ball (12.79 meters) to the height of the fence (5 feet ≈ 1.52 meters), we can determine that the ball does make it over the fence.
To determine if the ball makes it over the fence, we need to find the maximum height the ball reaches during its flight.
The initial velocity of the ball is given as 22 m/s at 45 degrees to the horizontal. We can break down this velocity into its vertical and horizontal components.
The vertical component can be found by multiplying the initial velocity by the sine of the launch angle:
Vertical component (V_y) = 22 m/s × sin(45°) = 22 m/s × (√2/2) ≈ 15.56 m/s
The horizontal component can be found by multiplying the initial velocity by the cosine of the launch angle:
Horizontal component (V_x) = 22 m/s × cos(45°) = 22 m/s × (√2/2) ≈ 15.56 m/s
The time it takes for the ball to reach its maximum height can be found using the vertical component of velocity:
Using the kinematic equation:
V_f = V_i + at
Since the ball reaches its maximum height, the final vertical velocity (V_f) will be zero. We can assume the acceleration due to gravity (a) is -9.8 m/s² (negative because the ball is moving upwards). The initial vertical velocity (V_i) is 15.56 m/s:
0 = 15.56 m/s - 9.8 m/s² × t
Solving for t:
t = 15.56 m/s / 9.8 m/s² ≈ 1.59 s
Next, we can find the maximum height (h) by using the formula:
h = V_i × t - (1/2) × a × t²
h = 15.56 m/s × 1.59 s - (1/2) × 9.8 m/s² × (1.59 s)²
h ≈ 12.35 meters
Since the maximum height of the ball is 12.35 meters, which is higher than the 5-foot (approximately 1.524 meters) tall fence, the ball does make it over the fence.