Chapter 04, Problem 116 Your answer is partially correct. Try again. An elevator

Question

Chapter 04, Problem 116 Your answer is partially correct. Try again. An elevator without a ceiling is ascending with a constant speed of 12 m/s. A boy on the elevator shoots a ball directly upward, from a height of 2.1 m above the elevator floor, just as the elevator floor is 25 m above the ground. The initial speed of the ball with respect to the elevator is 16.4 m/s. (a) What maximum height above the ground does the ball reach? (b) How long does the ball take to return to the elevator floor? (a) Numbef T68.25 UnitsT m (b) Number T3.3

Explanation / Answer

b)For the second part we have to calculate first the height attained by the ball when its thrown which can be found out by the formula v2=2gh by putting v=16.4m/s. On solving we get

h=v2/2g

h=16.42/2*9.8=13.72m/s

but the ball falls to the floor so we have to add the 2.1 m to it , so on adding we get 15.82m

now by using the formula

h=1/2gt2 we can find out the total time taken by the ball to reach the floor

now t1=sqrt(2*13.72/9.8), t2=sqrt(2*15.82/9.8), on adding both we get the total time

t1+t2=sqrt(2*13.72/9.8)+sqrt(2*15.82/9.8)=3.45 seconds

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.