Question

Use \(a(t)=-32\) feet per second per second as the acceleration due to gravity. A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet per second. How high will the ball go?

Short answer

The maximum height that the ball will reach is 56.25 feet.

Step-by-step solution

Recognise the given parameters

The problem provides these parameters: initial velocity \(v_0 = 60\) feet per second, initial height \(h_0 = 6\) feet, and acceleration \(a = -32\) feet per second per second (indicating downwards acceleration). The unknown parameter is the maximum height \(h_{max}\) the ball will attain.

Apply the principle of motion

The maximum height is attained when the velocity of the ball becomes zero, i.e., \(v = 0\). This is due to the fact that the ball changes its direction of motion at the maximum height. We can use the equation \(v^2 = v_0^2 + 2a(h - h_0)\), which is derived from the basic principles of motion under constant acceleration. At \(h_{max}\), the velocity \(v\) is zero. So, substituting the given values, we get \(0 = 60^2 + 2*-32*(h_{max} - 6)\).

Solve the equation

By solving the equation, we find \(h_{max} = 60^2/(2*-32) + 6 = 56.25\) feet.