Question

a bottle rocket is launched straight upwards with an initial velocity of 100 m/s. its height after t seconds is given by ( h(t)=100t - 5t^{2} ). find the velocity of the bottle rocket after 1 seconds. m/s find the velocity of the bottle rocket at ( t = a ) seconds. m/s when will the bottle rocket hit the ground? seconds what is the velocity of the bottle rocket as it hits the ground? m/s question help: message instructor

Explanation:

Step1: Find the derivative of \(h(t)\)

The velocity function \(v(t)\) is the derivative of the height function \(h(t)\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(h(t)=100t-5t^{2}\), we have \(v(t)=h^\prime(t)=(100t)^\prime-(5t^{2})^\prime\).
\(v(t)=100 - 10t\)

Step2: Calculate \(v(1)\)

Substitute \(t = 1\) into \(v(t)\).
\(v(1)=100-10\times1=90\)

Step3: Calculate \(v(a)\)

Substitute \(t=a\) into \(v(t)\).
\(v(a)=100 - 10a\)

Step4: Find when the rocket hits the ground

Set \(h(t)=0\), so \(100t-5t^{2}=0\). Factor out \(5t\): \(5t(20 - t)=0\).
We get \(t = 0\) (launch time) or \(t = 20\) (hit - ground time)

Step5: Calculate \(v(20)\)

Substitute \(t = 20\) into \(v(t)\).
\(v(20)=100-10\times20=- 100\)

Answer:

\(90\) m/s
\(100 - 10a\) m/s
\(20\) seconds
\(-100\) m/s