Final answer:
To find the maximum height of a thrown ball, we use the given quadratic function and the known values to calculate the vertex of the corresponding parabola. Plugging in the initial velocity, launch angle, and initial height into the function, and then using the time at the vertex, we find the max height is approximately 156.25 feet. Therefore, the closest answer is 150 feet.
Step-by-step explanation:
To determine the maximum height of a ball thrown at an angle with an initial velocity and initial height, we use the quadratic function h(t) = -16t2 + (v0sinA)t + h0, where:
- v0 is the initial velocity
- A is the angle of launch
- h0 is the initial height
- t is the time in seconds
For a ball thrown at a 30° angle, with an initial velocity of 100 ft/sec, and an initial height of 100 ft, we plug these values into the function:
h(t) = -16t2 + (100*sin30)t + 100
Since sin30 = 0.5, the equation simplifies to:
h(t) = -16t2 + 50t + 100
The maximum height is found at the vertex of the parabola represented by this quadratic equation. The time at which the maximum height occurs can be found by using the formula t = -b/2a, where a is the coefficient of t2 and b is the coefficient of t.
t = -50 / (2*-16) = 50 / 32
Once we have the time, plug it back into the function to find the maximum height:
h(50/32) = -16(50/32)2 + 50*(50/32) + 100
Solving for h(50/32), we find that the maximum height is approximately 156.25 feet. Hence, the closest answer from the options given is 150 feet (Option B).