Question

An object of mass 0.50 kg is released from the top of a building of height 8 m. The object experiences a horizontal constant force of 1.9 N due to a wind blowing parallel to the face of the building. (a) Find the time it takes for the object to strike the ground. (b) What is the magnitude of the acceleration of the object? (c) Through what horizontal distance does the object move before it hits the ground?

Answer 1

To solve this problem with the given elements we will apply the linear motion kinematic equations. We will start by calculating the time taken, with the vertical displacement data. Subsequently, with the components of the acceleration, we will obtain the magnitude of the total acceleration, to finally obtain the horizontal displacement with the data already found.

PART A) From vertical movement we know that the acceleration is equivalent to gravity and the displacement is 8m so the time taken to carry out the route would be

$h = \frac{1}{2} gt^2$

Here,

$h = 8 m$

$g = 9.8 m/s^2$

Replacing,

$8 = 0.5 * 9.8 * t^2$

$t = 1.277 sec$

PART B) Now, Magnitude of acceleration

$a = \sqrt{a_x^2 + a_y^2}$

$a_x = \frac{1.9}{0.5} = 3.8 m/s^2$

$a_y = g = 9.8 m/s^2$

Thus, magnitude of net acceleration

$a = \sqrt{3.82^2 + 9.8^2}= 10.51 m/s^2$

PART C) Finally the displacement along horizontal direction is:

$s =v_0 t + \frac{1}{2} a t^2$

$s = 0 + \frac{1}{2} (3.8)(1.277)^2$

$s = 3.098 m$

Therefore the distance traveled along the horizontal direction before it hits the ground is 3.098m