Find the height of a ball after 3 seconds, if the ball is hit with an initial velocity of 90 feet per second at an angle of 60 degrees from an initial height of 5 feet.
Vo = 90Ft/s[60o]
Xo = 90*cos60 = 45 Ft/s
Yo = 90*sin60 = 77.9 Ft/s
ho = 5 Ft.
g = -32 Ft/s^2
t = 3 s.
h = ho + Yo*t + 0.5g*t^2
To find the height of the ball after 3 seconds, we can use the equations of motion for a projectile.
Step 1: Resolve the initial velocity into its horizontal and vertical components.
The vertical component of the initial velocity can be found using trigonometry:
Vertical component = Initial velocity * sin(angle)
Vertical component = 90 ft/s * sin(60 degrees) ≈ 77.94 ft/s
Step 2: Calculate the time it takes for the ball to reach its maximum height.
The time it takes for a projectile to reach its maximum height can be found using the equation:
Time to max height = (Vertical component of initial velocity) / (acceleration due to gravity)
Time to max height = 77.94 ft/s / 32.17 ft/s² ≈ 2.42 seconds
Step 3: Calculate the maximum height reached by the ball.
The maximum height can be found using the equation:
Max height = Initial height + (Vertical component of initial velocity * time to max height) - (0.5 * acceleration due to gravity * (time to max height)²)
Max height = 5 ft + (77.94 ft/s * 2.42 s) - (0.5 * 32.17 ft/s² * (2.42 s)²)
Max height ≈ 153.53 ft
Step 4: Calculate the time it takes for the ball to reach the ground.
The total time of flight can be found using the equation:
Time of flight = 2 * (Vertical component of initial velocity) / (acceleration due to gravity)
Time of flight = 2 * (77.94 ft/s) / (32.17 ft/s²) ≈ 4.83 seconds
Step 5: Calculate the height of the ball after 3 seconds.
Since 3 seconds is within the time of flight, we can use the equation for the vertical position of a projectile at any given time:
Vertical position = Initial height + (Vertical component of initial velocity * time) - (0.5 * acceleration due to gravity * time²)
Vertical position = 5 ft + (77.94 ft/s * 3 s) - (0.5 * 32.17 ft/s² * (3 s)²)
Vertical position ≈ 137.85 ft
Therefore, the height of the ball after 3 seconds is approximately 137.85 feet.
To find the height of the ball after 3 seconds, we need to break down the problem into its horizontal and vertical components.
First, let's find the initial velocity of the ball in the horizontal and vertical directions using trigonometry. The initial velocity in the horizontal direction is given by:
Vx = initial velocity * cos(angle)
Vx = 90 ft/sec * cos(60 degrees)
Vx = 90 ft/sec * 0.5
Vx = 45 ft/sec
The initial velocity in the vertical direction is given by:
Vy = initial velocity * sin(angle)
Vy = 90 ft/sec * sin(60 degrees)
Vy = 90 ft/sec * (√3/2)
Vy = 90/2 * √3 ft/sec
Vy = 45 * √3 ft/sec
Now, we can find the height of the ball after 3 seconds using the following formula for vertical motion:
h = Vy * t - (1/2) * g * t^2
where h is the height, Vy is the initial vertical velocity, t is the time in seconds, and g is the acceleration due to gravity (approximately 32.2 ft/sec^2).
Substituting the given values:
h = (45 * √3 ft/sec) * 3 sec - (1/2) * 32.2 ft/sec^2 * (3 sec)^2
h = (135 * √3 ft) - (1/2) * 32.2 ft/sec^2 * 9 sec^2
h = 135 * √3 ft - 144.9 ft/sec^2 * 9 sec^2
h = 135 * √3 ft - 144.9 ft/sec^2 * 81 sec^2
h = 135 * √3 ft - 11738.9 ft * sec^2
Calculating this expression, we find:
h ≈ 232.62 ft
Therefore, the height of the ball after 3 seconds is approximately 232.62 feet.