Questions: The height in feet of a projectile with an initial velocity of 64 feet per second and an initial height of 80 feet is a function time in seconds given by h(t)=-16 t^2+64 t+80 . (a) Find the maximum height of the projectile. ft (b) Find the time t when the projectile achieves its maximum height. t= sec (c) Find the time t when the projectile has a height of 0 feet. t= sec

Solution Steps

The height function \( h(t) = -16t^2 + 64t + 80 \) is a quadratic function, which is a parabola opening downwards (since the coefficient of \( t^2 \) is negative). The maximum height is at the vertex of the parabola.

The time \( t \) at which the maximum height occurs can be found using the vertex formula for a parabola \( t = -\frac{b}{2a} \), where \( a = -16 \) and \( b = 64 \).

\[ t = -\frac{64}{2 \times (-16)} = \frac{64}{32} = 2 \text{ seconds} \]

Substitute \( t = 2 \) back into the height function to find the maximum height:

\[ h(2) = -16(2)^2 + 64(2) + 80 = -64 + 128 + 80 = 144 \text{ feet} \]

From the calculation in Step 1, the time \( t \) when the projectile achieves its maximum height is 2 seconds.

To find when the projectile hits the ground, set \( h(t) = 0 \) and solve for \( t \):

\[ -16t^2 + 64t + 80 = 0 \]

This is a quadratic equation, which can be solved using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -16 \), \( b = 64 \), and \( c = 80 \).

\[ t = \frac{-64 \pm \sqrt{64^2 - 4 \times (-16) \times 80}}{2 \times (-16)} \]

\[ t = \frac{-64 \pm \sqrt{4096 + 5120}}{-32} \]

\[ t = \frac{-64 \pm \sqrt{9216}}{-32} \]

\[ t = \frac{-64 \pm 96}{-32} \]

This gives two solutions:

\[ t_1 = \frac{-64 + 96}{-32} = \frac{32}{-32} = -1 \text{ (not physically meaningful)} \]

\[ t_2 = \frac{-64 - 96}{-32} = \frac{-160}{-32} = 5 \text{ seconds} \]

Final Answer