A golf ball is dropped from rest from a height of 8.10 m. It hits the pavement,

Question

A golf ball is dropped from rest from a height of 8.10 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.50 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.

Explanation / Answer

sol. The first drop took T(1) where ==> (1/2)g{ T(1)^2 } = 8.10 m and g = 9.8 m/s^2 It then bounces having lost some energy but rebounding with a velocity V taking it to a height of 5.60m in a time T(2) where ( by symmetry ) ==> 5.6 = (1/2)g{T(2)^2}. On the way down again, however, the bag falls only 5.60-1.50= 4.10 m before the boy catches it after another T(3). Again, == > (1/2)g{ T(3)^2 } = 4.10 m. Solve for T(1), T(2) and T(3) and sum them for your answe

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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