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? d. 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 \( v(t)=-1.6 \mathrm{t}+48 \) meters per second. b. It takes the object (Simplify your answer.) meters

The height function you've been given, \( s(t) = -0.8t^2 + 48t \), is a parabolic equation that describes the motion of the object under the influence of gravity. To find the velocity after \( t \) seconds, you can take the derivative of the height function: \( v(t) = s'(t) = -1.6t + 48 \) meters per second. To determine when the object reaches its highest point, set the velocity \( v(t) \) to zero: \[ -1.6t + 48 = 0 \] Solving for \( t \) gives: \[ t = \frac{48}{1.6} = 30 \ \text{seconds} \] This is when the object reaches its apex. For the height at this peak, substitute \( t = 30 \) back into the height equation: \[ s(30) = -0.8(30^2) + 48(30) = -720 + 1440 = 720 \ \text{meters} \] Thus, the object reaches 720 meters at the highest point. To find when the object strikes the ground, set \( s(t) = 0 \): \[ -0.8t^2 + 48t = 0 \] Factoring out \( t \): \[ t(-0.8t + 48) = 0 \] This gives \( t = 0 \) (launch time) and \( t = 60 \ \text{seconds} \) (when it hits the ground). Finally, to calculate the velocity as it strikes the ground, substitute \( t = 60 \): \[ v(60) = -1.6(60) + 48 = -96 + 48 = -48 \ \text{m/s} \] Thus, it strikes the ground at a speed of 48 meters per second downward. To analyze intervals of increasing speed, observe that speed is increasing when both velocity and acceleration are either positive or negative together. Since the acceleration due to gravity is constant and negative, the object is always slowing while going up, but after it starts descending, it gains speed until it strikes the ground. Thus, speed is increasing for \( t > 30 \) seconds until it strikes the ground at \( t = 60 \) seconds. Here’s a quick wrap-up: - Velocity: \( v(t) = -1.6t + 48 \) m/s - Highest point time: \( 30 \, \text{seconds} \) - Height at the highest point: \( 720 \, \text{meters} \) - Ground strike time: \( 60 \, \text{seconds} \) - Ground strike velocity: \( -48 \, \text{m/s} \) (downwards) - Speed is increasing from \( 30 \, \text{seconds} \) to \( 60 \, \text{seconds} \). Let me know if you have any more questions or need further clarification!