A ball is tossed up in the air at an angle of 70 degrees with the horizontal and with an initial velocity of 36 ft/sec. What is the position of the ball 1 second after it is released? When will the ball hit the ground? What is the maximum height the ball will reach? How far in a horizontal direction will the ball travel?
the velocity has two components:
x: 36 cos70° = 12.31
y: 36 sin70° = 33.83
now recall that the x-speed is constant, and the height
y = 33.8t - 16t^2
now solve for the various requirements
To find the position of the ball 1 second after it is released, we need to break down the initial velocity into its horizontal and vertical components.
The initial velocity can be split into two components using the given angle of 70 degrees with the horizontal. The horizontal component is given by v₀x = v₀ * cos(θ), and the vertical component is given by v₀y = v₀ * sin(θ).
Given that the initial velocity, v₀, is 36 ft/sec, and the angle, θ, is 70 degrees, we can calculate the horizontal and vertical components as follows:
v₀x = 36 * cos(70)
v₀y = 36 * sin(70)
The horizontal component of the velocity remains constant throughout the motion, so after 1 second, the horizontal position of the ball will be:
x = v₀x * t
Substituting the values, we get:
x = (36 * cos(70)) * 1
To find the vertical position of the ball after 1 second, we need to use the equation of motion for the vertical direction:
y = y₀ + v₀y * t - (1/2) * g * t²
Here, y₀ represents the initial vertical position, and g is the acceleration due to gravity. Since the ball is released at ground level, y₀ is 0. The value of g is approximately -32 ft/sec² (negative because it acts in the opposite direction to the upward motion).
Substituting the values, we get:
y = 0 + (36 * sin(70)) * 1 - (1/2) * (-32) * (1)^2
Now, let's calculate the position after 1 second:
x = (36 * cos(70)) * 1 ≈ 11.21 ft
y = (36 * sin(70)) - 16 ≈ 31.99 ft
Therefore, the position of the ball 1 second after it is released is approximately (11.21 ft, 31.99 ft).
Next, let's find when the ball will hit the ground. Since the ball was initially at ground level, the time it takes to hit the ground can be calculated by setting y = 0 in the equation:
0 = (36 * sin(70)) * t - (1/2) * (-32) * t^2
Simplifying this equation, we get:
0 = 18t - 16t²
Solving for t, we get two possible values, t = 0 and t ≈ 1.13 seconds. The first value of t = 0 represents the initial time, so the ball will hit the ground approximately 1.13 seconds after it was released.
To find the maximum height the ball will reach, we need to find the maximum point of its vertical motion. The maximum height occurs when the vertical velocity component becomes zero (v_y = 0). We can calculate this using the equation:
v_y = v₀y - g * t
Setting v_y = 0, we can solve for t:
0 = (36 * sin(70)) - 32t
Simplifying and solving for t, we get:
t ≈ (36 * sin(70)) / 32 ≈ 1.20 seconds
Finally, substituting this value of t into the equation for y, we can find the maximum height:
y = (36 * sin(70)) * 1.20 - (1/2) * (-32) * (1.20)^2
Simplifying this equation, we get:
y ≈ 37.16 ft
Therefore, the ball will reach a maximum height of approximately 37.16 ft.
Lastly, to determine how far the ball will travel in the horizontal direction, we can use the equation:
x = v₀x * t
Substituting the given values, we get:
x = (36 * cos(70)) * 1.13
Calculating this, we find:
x ≈ 10.72 ft
Therefore, the ball will travel approximately 10.72 ft in the horizontal direction.