Questions: A ball is thrown straight up in the air at a velocity of 100 feet per second. The height of the ball at t seconds is given by the formula: h=100 t-5 t^2. Find the maximum height of the ball and when it will land. Maximum height: ft Land after: seconds

Solution Steps

To find the time \( t \) at which the ball reaches its maximum height, we need to find the vertex of the parabolic function \( h(t) = 100t - 5t^{2} \). The time \( t \) at which the maximum height occurs is given by: \[ t = \frac{-b}{2a} \] where \( a = -5 \) and \( b = 100 \). Substituting these values: \[ t = \frac{-100}{2(-5)} = \frac{-100}{-10} = 10 \text{ seconds}. \]

Substitute \( t = 10 \) seconds into the height formula \( h(t) = 100t - 5t^{2} \) to find the maximum height: \[ h(10) = 100(10) - 5(10)^{2} = 1000 - 5(100) = 1000 - 500 = 500 \text{ feet}. \]

The ball will land when its height \( h(t) = 0 \). Set the height equation equal to zero and solve for \( t \): \[ 0 = 100t - 5t^{2}. \] Factor out \( t \): \[ 0 = t(100 - 5t). \] This gives two solutions: \[ t = 0 \quad \text{or} \quad 100 - 5t = 0. \] Solving the second equation: \[ 100 - 5t = 0 \implies 5t = 100 \implies t = 20 \text{ seconds}. \] Since \( t = 0 \) corresponds to the initial time, the ball will land after \( t = 20 \) seconds.

Final Answer