Question

An object is thrown vertically upward with an initial velocity of 75ft/sec from a building 40ft above the ground. (a = 32ft/sec²)

1. Determine the equation of the distance s from the ground at a given time t in seconds.

2. Calculate the distance of the object from the ground after 2 seconds.

Answer 1

1. Let y-axis is directed upward from the ground. Then $y(t)$ will be the distance from the ground at a given time $t$ in seconds.

$$a=-32\ ft/s^2, v_0=75\ ft/s, y_0=40\ ft$$

$$v(t)=\int a dt=v_0+at$$

$$y(t)=\int vdt=y_0+v_0t+\dfrac{at^2}{2}$$

The equation of the motion along y-axis is

$$y(t)=y_0+v_0t+\dfrac{at^2}{2}, t\geq0$$

Substitute

$$y(t)=40+75t-\dfrac{32t^2}{2}, ft$$

$$y(t)=40+75t-16t^2, ft$$

2.

$$y(2)=40+75(2)-16(2)^2=126(ft)$$

The distance of the object from the ground after 2 seconds is 126 ft.