Skip to content
Exercise 2.3 · Q8

Q.A model rocket is launched from the ground. The height hh reached by the rocket after tt seconds from lift off is given by h(t)=−5t2+100t, 0≤t≤20h(t)=-5t^2+100t,\ 0\le t\le20. At what time is the rocket 495 feet above the ground?

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
15% · 19/128 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Set −5t2+100t=495-5t^2+100t=495.

Step 2. Rearrange: −5t2+100t−495=0-5t^2+100t-495=0. Divide by −5-5: t2−20t+99=0t^2-20t+99=0.

Step 3. Factor: (t−9)(t−11)=0(t-9)(t-11)=0, so t=9t=9 or t=11t=11. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.