Question

Dan is training his puppy, Lola, to sit. Dan holds a treat 5 feet above the ground until Lola

sits. As soon as she does, he drops the treat, and Lola catches it in her mouth 1.5 feet above
the ground.
To the nearest tenth of a second, how long does the treat fall before Lola catches it?
Hint: Use the formula h = -16t² + s.

Answer 1

t = 0.5 seconds

Explanation:

Given formula:

`h=-16t^2+s`

where:

• h = height above the ground (in feet)
• t = time after dropping the treat (in seconds)

Dan drops the treat 5 feet above the ground.

Therefore, when t = 0, h = 5 ft.

To find the value of s, substitute these values into the given formula:

`\implies -16(0)^2+s=5`

`\implies s=5`

Therefore:

`h=-16t^2+5`

To calculate how long the treat falls before Lola catches it, substitute h = 1.5 into the formula:

`\implies -16t^2+5=1.5`

`\implies -16t^2=-3.5`

`\implies t^2=0.21875`

`\implies t=√(0.21875)`

`\implies t=\pm0.467707173...`

`\implies t=\pm0.5\; \; \sf (nearest\;tenth)`

As time is positive, t = 0.5 seconds (nearest tenth) only.

Therefore, it took 0.5 seconds for Lola to catch the treat after it fell.