Answer:
t = 3.01 s
Explanation:
In order to calculate how long it takes to the object to fall to the ground, you use the following formula, for the calculation of the height:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
yo: height of the building = 105 m
vo: initial velocity of the body = 20m/s
g: gravitational acceleration = 9.8m/s^2
t: time = ?
To find the time t, you take into account that when the body arrives to the ground the height is zero, that is, y = 0.
You replace the values of all parameters in the equation (1), and you obtain a quadratic polynomial for t:
[tex]0=105-20t-\frac{1}{2}(9.8)t^2\\\\0=-4.9t^2-20t+105[/tex]
Next, you use the quadratic formula to get the roots of the polynomial:
[tex]t_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
a = -4.9
b = -20
c = 105
[tex]t_{1,2}=\frac{-(-20)\pm\sqrt{(-20)^2-4(-4.9)(105)}}{2(-4.9)}\\\\t_1=3.01s\\\\t_2=-7.09s[/tex]
You choose the positive value t1, because it has physical meanning.
Hence, the body takes 3.01s to arrive to the ground
