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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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Verified Answer :

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.

JeanaShupp

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.]

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