If ball is to be thrown vertically upwards, the ball reaches a height of \( y=-5 t^{2}+20 t+5 \) feet after \( t \) seconds. (2dp) [7] a. What maximum height will the ball be able to reach? b. What is the height of the ball when the velocity is \( 6 \mathrm{ft} / \mathrm{s} \) ?

To find the maximum height, we can analyze the quadratic function given by \( y(t) = -5t^2 + 20t + 5 \). The maximum height occurs at the vertex of the parabola described by this equation. We can find the time at which the vertex occurs using the formula \( t = -\frac{b}{2a} \), where \( a = -5 \) and \( b = 20 \). Calculating: \[ t = -\frac{20}{2 \times -5} = 2 \text{ seconds} \] Substituting \( t = 2 \) back into the height equation: \[ y(2) = -5(2)^2 + 20(2) + 5 = -20 + 40 + 5 = 25 \text{ feet} \] Hence, the maximum height the ball can reach is **25 feet**. To find the height when the velocity is \( 6 \, \text{ft/s} \), we first need to determine the velocity function \( v(t) \), which is the derivative of the height function: \[ v(t) = \frac{dy}{dt} = -10t + 20 \] Setting \( v(t) = 6 \): \[ -10t + 20 = 6 \implies -10t = 6 - 20 \implies -10t = -14 \implies t = 1.4 \text{ seconds} \] Now, substituting \( t = 1.4 \) into the height equation: \[ y(1.4) = -5(1.4)^2 + 20(1.4) + 5 \approx -5(1.96) + 28 + 5 \approx -9.8 + 28 + 5 \approx 23.2 \text{ feet} \] Thus, the height of the ball when the velocity is \( 6 \text{ ft/s} \) is approximately **23.20 feet**.