danielmassey7
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Dave throws sweets upwards at a speed of 8ms^-1 from a first floor window to his sister Carol who is standing at a second floor window 3m directly above Dave. Carol misses catching the sweet as it passes her hand on the way up, but catches it on the way down. Calculate the time it takes for Carol to catch the sweet after Dave has thrown it.

Sagot :

batolisis

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

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