a bottle rocket is launched straight upwards with an initial velocity of 100 m/s. its height after t seconds…

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

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

Answer

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