Questions: A projectile is launched upward with a velocity of 256 feet per second from the top of a 60 foot stage. What is the maximum height attained by the projectile?

Solution Steps

The problem involves a projectile launched upward with an initial velocity from a certain height. We need to find the maximum height attained by the projectile. The initial velocity is given as 256 feet per second, and the initial height is 60 feet.

The height \( h(t) \) of the projectile at time \( t \) can be described by the kinematic equation:

\[ h(t) = h_0 + v_0 t - \frac{1}{2} g t^2 \]

where:

• \( h_0 = 60 \) feet is the initial height,
• \( v_0 = 256 \) feet/second is the initial velocity,
• \( g = 32 \) feet/second\(^2\) is the acceleration due to gravity.

The maximum height is reached when the velocity is zero. The velocity \( v(t) \) is given by the derivative of the height function:

\[ v(t) = v_0 - g t \]

Setting \( v(t) = 0 \) to find the time \( t \) at maximum height:

\[ 256 - 32t = 0 \]

Solving for \( t \):

\[ t = \frac{256}{32} = 8 \text{ seconds} \]

Substitute \( t = 8 \) seconds back into the height equation to find the maximum height:

\[ h(8) = 60 + 256 \times 8 - \frac{1}{2} \times 32 \times 8^2 \]

Calculate each term:

• \( 256 \times 8 = 2048 \)
• \( \frac{1}{2} \times 32 \times 8^2 = 1024 \)

Thus, the maximum height is:

\[ h(8) = 60 + 2048 - 1024 = 1084 \text{ feet} \]

Final Answer