Answer:
Option b is correct: 4.1 s
Explanation:
Vertical Launch
An object launched thrown vertically upward where air resistance is negligible, reaches its maximum height in a time t, given by the equation:
[tex]\displaystyle t=\frac{v_o}{g}\qquad\qquad[1][/tex]
Where vo is the initial speed and g is the acceleration of gravity g=9.8 [tex]m/s^2[/tex].
Once the object reaches that point, it starts a free-fall motion, whose speed is (downward) given by:
[tex]v_f=g.t\qquad\qquad[2][/tex]
The object considered in the question is thrown with vo=25 m/s. The time taken to reach the maximum height is given by [1]:
[tex]\displaystyle t=\frac{25}{9.8}=2.551\ sec[/tex]
The object starts its falling motion and at some time, it has a speed of vf=15 m/s. Let's find the time by solving [2] for t:
[tex]\displaystyle t=\frac{15}{9.8}=1.531\ sec[/tex]
The total time taken by the object to go up and down is
[tex]t_t=2.551\ s+1.531\ s=4.081\ s[/tex]
a. This option is incorrect because it's far away from the answer.
d. This option is incorrect because it's far away from the answer.
b. This option is correct because it's a good approximation to the calculated answer.
e. This option is incorrect because it's far away from the answer.
c. This option is incorrect because it's far away from the answer.
