Answer to Question #86957 in Mechanics | Relativity for Jonathan

Question

An object is thrown upward from point A with a velocity of 40ms-1 calculate the time it takes to return to point A

Answer 1

Let's first find the time an object takes to reach the maximum height from the kinematic equation:

$$v=v_0 - gt,$$

here, $v_0$ is initial velocity of the object, $v = 0$ is the velocity of the object at the maximum height, $g$ is the acceleration due to gravity and $t$ is the time the object takes to reach the maximum height.

Then, we get:

$$t = \dfrac{v_0}{g} = \dfrac{40 \dfrac{m}{s}}{9.8 \dfrac{m}{s^2}} = 4.1 s.$$

Finally, we can find the total time the object takes to return to point A:

$$t_{tot} = 2t =2 \cdot 4.1 s = 8.2 s.$$

Answer:

$t_{tot} = 8.2 s.$