Question

4. An object is thrown vertically upward with a speed of 25 m/s. How much time passes before it

Answer 1

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:

$\displaystyle t=\frac{v_o}{g}\qquad\qquad[1]$

Where vo is the initial speed and g is the acceleration of gravity g=9.8 $m/s^2$.

Once the object reaches that point, it starts a free-fall motion, whose speed is (downward) given by:

$v_f=g.t\qquad\qquad[2]$

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]:

$\displaystyle t=\frac{25}{9.8}=2.551\ sec$

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:

$\displaystyle t=\frac{15}{9.8}=1.531\ sec$

The total time taken by the object to go up and down is

$t_t=2.551\ s+1.531\ s=4.081\ s$

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.