If a ball is thrown in the air with an initial height of 3 feet, and if the ball remains in the air for 4 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration due to gravity on Earth is \( -32 \mathrm{ft} / \mathrm{sec}^{2} \). [? ] feet

To determine the maximum height reached by the ball, we'll use the kinematic equation for vertical motion: \[ h(t) = h_0 + v_0 t - \frac{1}{2} g t^2 \] where: - \( h(t) \) is the height at time \( t \), - \( h_0 = 3 \) (initial height), - \( v_0 \) is the initial velocity, - \( g = 32 \, \text{ft/sec}^2 \) (the acceleration due to gravity). First, we need to find the initial velocity. The ball reaches its highest point at half of the total time, so at \( t = 2 \) seconds. We can set the velocity \( v(t) \) at the highest point to zero: \[ v(t) = v_0 - gt \] Setting \( v(2) = 0: \] \[ 0 = v_0 - 32 \cdot 2 \] \[ v_0 = 64 \, \text{ft/sec} \] Now we can calculate the maximum height at \( t = 2 \): \[ h(2) = 3 + 64 \cdot 2 - \frac{1}{2} \cdot 32 \cdot (2^2) \] \[ h(2) = 3 + 128 - 64 \] \[ h(2) = 67 \, \text{feet} \] Thus, the maximum height the ball reached is approximately **67 feet**. Do keep in mind, if the ball is in the air for 4 seconds, it means it returns to the initial height after 4 seconds, confirming our math was spot on!