Questions: If a ball is thrown in the air with an initial height of 3 feet, and if the ball remains in the air for 3 seconds, then accurate to the nearest foot, how high did it go? Remember, the acceleration is a(t)=-32 ft / sec^2 v(t)=-32 t+47 s(t)=-16 t^2+47 t+3 Peak height =s(1.47) s(1.47)=[?] feet

Solution Steps

The problem involves finding the maximum height reached by a ball thrown into the air. The ball's motion is described by the position function \( s(t) = -16t^2 + 47t + 3 \), where \( t \) is time in seconds. The acceleration due to gravity is given as \( a(t) = -32 \, \text{ft/sec}^2 \).

The maximum height is reached when the velocity \( v(t) \) is zero. The velocity function is given by \( v(t) = -32t + 47 \). Set \( v(t) = 0 \) to find the time \( t \) when the ball reaches its peak height: \[ -32t + 47 = 0 \] \[ t = \frac{47}{32} \approx 1.47 \, \text{seconds} \]

Substitute \( t = 1.47 \) into the position function \( s(t) \) to find the maximum height: \[ s(1.47) = -16(1.47)^2 + 47(1.47) + 3 \] Calculate \( s(1.47) \) to find the height in feet.

Final Answer