It would take [tex]1 second[/tex] for Carol to catch the sweet after Dave threw it.
Let
[tex]h=\text{maximum height}\\y=\text{distance from maximum height to Carol}\\t_h=\text{time taken to reach maximum height}\\t_y=\text{time taken to fall from maximum height to Carol}[/tex]
we want to calculate the value [tex]t_h+t_y[/tex]
First, we have to calculate [tex]t_h[/tex] using the formula
[tex]v=u+at[/tex]
using [tex]v=0[/tex], [tex]u=8[/tex], [tex]a=-10[/tex] (the sweets are decelerating due to gravity!)
[tex]0=8-(10)t_h\\10t_h=8\\t_h=\frac{8}{10}=0.8seconds[/tex]
To calculate [tex]t_y[/tex], first calculate [tex]h[/tex], then [tex]y[/tex]. Using the formula
[tex]h=ut+\frac{1}{2}at^2[/tex]
and the values [tex]u=8[/tex], [tex]t=t_h=0.8[/tex], [tex]a=-10[/tex] (again, the sweets are decelerating to get to maximum height)
[tex]h=(8)(0.8)+\frac{1}{2}(-10)(0.8)^2\\\\=6.4-3.2\\=3.2m[/tex]
since, Carol is standing [tex]3m[/tex] above Dave, we have the relationship
[tex]h=y+3[/tex]
so that
[tex]y=h-3\\=3.2-3\\=0.2m[/tex]
we can now calculate [tex]t_y[/tex];
[tex]y=ut_y+\frac{1}{2}gt_y^2[/tex]
taking [tex]y=0.2[/tex], [tex]u=0[/tex], [tex]g=10[/tex] (the sweets are falling, so they are now accelerating)
[tex]0.2=(0)t_y+\frac{1}{2}(10)t_y^2\\\\0.2=5t_y^2\\t_y^2=0.04\\\implies t_y=0.2seconds[/tex]
So it would take
[tex]t_h+t_y=0.8+0.2=1.0seconds[/tex]
for Carol to catch the sweet after Dave threw it from the first floor.
Learn more here: https://brainly.com/question/84352
