Question

A construction worker drops a hammer from a height of 25 m.
How long will it take the hammer to hit the ground?
Use the formula $h(t)=-4.9t^2+v_o+h_o$ , where $v_o$ is the initial velocity and $h_o$ is the initial height.
Round to the nearest tenth of a second. Enter your answer in the box.
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Answer 1

Answer:

Time at which the hammer reaches the ground is 2.3 seconds.

Step-by-step explanation:

We are given the formula as, $h(t)=-4.9t^2+v_{o}+h_{o}$.

It is known that, when height i.e. $h_{o}=25$ meter, then velocity $v_{o}=0$ m/sec.

So, we get the formula as, $h(t)=-4.9t^2+25$.

Now, when the hammer hits the ground, the height $h_{o}=0$ meter.

Thus, we have,

$h(t)=-4.9t^2+25$

⇒ $0=-4.9t^2+25$

⇒ $4.9t^2=25$

⇒ $t^2=5.1$

i.e. t= ±2.3 sec

Since, time cannot be negative.

So, t= 2.3 sec

Hence, the time at which the hammer reaches the ground is 2.3 seconds.

Answer 2

Answer: 2.3 seconds

Step-by-step explanation:

Given: A construction worker drops a hammer from a height of $h_{o}=25\ meter$.

Using formula $h(t)=-4.9t^2+v_{o}+h_{o}$ , where $v_o$ is the initial velocity and $h_o$ is the initial height.

Put t=0, the initial velocity = 0m/s in the beginning.

$h(t)=-4.9t^2+0+25=4.9t^2+25$

When the hammer hits the ground, then height $h_{o}=0\ meter$

$h(t)=-4.9t^2+25\\\\\Rightarrow 0=-4.9t^2+25\\\\\Rightarrow 4.9t^2=25\\\\\Rightarrow t^2=5.\\\\\Rightarrow\ t=\sqrt{5.1}=\pm2.2583179581\approx \pm2.3$

$\\\\\Rightarrow t=2.3\ seconds$  [Since time is a positive quantity.]