An object thrown vertically upward from the surface of a celestial body at a velocity of \( 48 \mathrm{~m} / \mathrm{s} \) reaches a height of \( \mathrm{s}=-0.8 \mathrm{t}^{2}+48 \mathrm{t} \) meters in t seconds. a. Determine the velocity v of the object after t seconds. b. When does the object reach its highest point? c. What is the height of the object at the highest point? d. When does the object strike the ground? e. With what velocity does the object strike the ground? f . On what intervals is the speed increasing? a. The object's velocity after \( t \) seconds is \( \mathrm{v}(\mathrm{t})=\square \)

To find the velocity of the object after \( t \) seconds, we will differentiate the height function \( s(t) = -0.8t^2 + 48t \) with respect to time \( t \). The velocity \( v(t) \) is given by: \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(-0.8t^2 + 48t) = -1.6t + 48 \] So, the object's velocity after \( t \) seconds is \( \mathrm{v}(t) = -1.6t + 48 \). Now, let's move on to when the object reaches its highest point. To find the time \( t \) at which the object reaches its highest point, we set the velocity \( v(t) \) to zero: \[ 0 = -1.6t + 48 \] Solving for \( t \): \[ 1.6t = 48 \quad \Rightarrow \quad t = \frac{48}{1.6} = 30 \, \text{seconds} \] The object reaches its highest point at \( t = 30 \) seconds. Next, we need to find the height of the object at this point by substituting \( t = 30 \) into the height equation: \[ s(30) = -0.8(30^2) + 48(30) = -0.8(900) + 1440 = -720 + 1440 = 720 \, \text{meters} \] So, the height of the object at the highest point is \( 720 \) meters. To find when the object strikes the ground, we set the height \( s(t) \) to zero: \[ -0.8t^2 + 48t = 0 \] Factoring gives: \[ t(-0.8t + 48) = 0 \] Thus, the solutions are \( t = 0 \) (when it was thrown) and \( -0.8t + 48 = 0 \) leads to: \[ 0.8t = 48 \quad \Rightarrow \quad t = \frac{48}{0.8} = 60 \, \text{seconds} \] The object strikes the ground at \( t = 60 \) seconds. Now let's calculate the velocity with which the object strikes the ground by substituting \( t = 60 \) into the velocity equation: \[ v(60) = -1.6(60) + 48 = -96 + 48 = -48 \, \text{m/s} \] So the object strikes the ground with a velocity of \( -48 \, \text{m/s}\) (the negative sign indicates a downward direction). For the intervals where the speed is increasing, we check when the velocity is increasing, which occurs when the derivative of the velocity is positive. The derivative of the velocity function \( v(t) = -1.6t + 48 \) is constant at \( -1.6 \), indicating that the velocity is always decreasing. Therefore, the speed, being the magnitude of velocity, is not increasing at any point during the object's flight. Thus, there are no intervals where the speed is increasing. Summarizing: a. The object's velocity after \( t \) seconds is \( v(t) = -1.6t + 48 \). b. The object reaches its highest point at \( t = 30 \) seconds. c. The height at the highest point is \( 720 \) meters. d. The object strikes the ground at \( t = 60 \) seconds. e. The object strikes the ground with a velocity of \( -48 \, \text{m/s} \). f. The speed is never increasing during its flight.