Physics Question

Question

A golf ball is dropped from rest from a height of 8.30 m. It hits the pavement, then bounces back up, rising just 5.40 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.

 

Help?!

Answer 1

I will set up the solution, but YOU do the calculations.  You have to calculate separately the duration of 3 different time intervals, and then add the 3 durations to get the total time the ball is in the air.

 

(1)

 

Drop from 8.30 m to ground:

 

No air resistance ⇒

 

h₁ = gt₁²/2

 

t₁ = √(2h₁/g) = √[2(8.30)/9.8] sec = ? sec

 

(2) 

 

Bounce from ground up to 5.40 m:

 

The time it takes to bounce from 0 to 5.40 m must equal the time it would take to fall from 5.40 m to 0.

 

h₂ = gt₂²/2

 

t₂ = √(2h₂/g) = √[2(5.40)/9.8] sec = ? sec

 

(3)

 

Fall from 5.40 m to 1.50 m:

 

No air resistance ⇒

 

Δy = gt₃²/2

 

t₃ = √(2Δy/g)

 

Δy = (5.40 - 1.50) m = ? m

g = 9.8 m/s²

 

Plug in the numbers and get t₃.

 

TOTAL TIME IN AIR:

 

t = t₁ + t₂ + t₃ = ?