Solve the problem.
-An object is fired vertically upward with an initial velocity v(0) = v 0 from an initial position s(0) = s 0 . Assuming no air resistance, the position of the object is governed by the differential equation is the acceleration due to gravity (in the downward direction). For the following values of v 0 and s 0 , find the position and velocity functions for all times at which the object is above the ground. Then find the time at which the highest point of the trajectory is reached and the height of the object at that time.
v 0 = 39.2 m/s , s 0 = 40 m

A) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t; Highest point of 118.4 m is reached at t = 5 s
B) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s
C) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 5 s
D) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s

Question

Solve the problem.
-An object is fired vertically upward with an initial velocity v(0) = v 0 from an initial position s(0) = s 0 . Assuming no air resistance, the position of the object is governed by the differential equation     is the acceleration due to gravity (in the downward direction). For the following values of v 0 and s 0 , find the position and velocity functions for all times at which the object is above the ground. Then find the time at which the highest point of the trajectory is reached and the height of the object at that time.
v 0 = 39.2 m/s , s 0 = 40 m

A) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t; Highest point of 118.4 m is reached at t = 5 s 
B) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s 
C) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 5 s 
D) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s

Quizplus · Accepted Answer

The Answer of Solve the problem.
-An object is fired vertically upward with an initial velocity v(0) = v 0 from an initial position s(0) = s 0 . Assuming no air resistance, the position of the object is governed by the differential equation     is the acceleration due to gravity (in the downward direction). For the following values of v 0 and s 0 , find the position and velocity functions for all times at which the object is above the ground. Then find the time at which the highest point of the trajectory is reached and the height of the object at that time.
v 0 = 39.2 m/s , s 0 = 40 m

A) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t; Highest point of 118.4 m is reached at t = 5 s 
B) s = -9.8 $ t^{2} $ + 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s 
C) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 5 s 
D) s = -4.9 $ t^{2} $+ 39.2t + 40, v = -9.8t + 39.2; Highest point of 118.4 m is reached at t = 4 s

Solve the Problem t2 t^{2} t2 + 392t + 40, V = -9

Solve the Problem $t^{2}$ + 392t + 40, V = -9