Question

A rubber ball is tossed straight up from a height of 10 feet with a velocity of 78 feet per second. The first time it hits the ground (y = 0), it rebounds with a velocity of 64 feet per second2 . The second time it hits the floor, it rebounds with a velocity of 48 feet per second. Before the first bounce 1. Find the function y = h1(t) for the height of the ball before its first bounce. 2. Find the time t1 when the ball hits the ground for the first time. The function h1(t) then describes the height of the ball over the time interval [0, t1] before the ball’s first bounce. 3. Find the velocity v1(t) of the ball over the interval [0, t1]. 4. Determine the time t when the ball reaches its first peak height. What is the height of the ball at this time? Between the first and second bounce 5. Find the function y = h2(t) for the height of the ball between the first and second bounce. 6. Find the time t2 when the ball hits the ground for the second time. The function h2(t) then describes the height of the ball over the time interval [t1, t2]. 7. Find the velocity v2(t) of the ball over the interval [t1, t2]. 8. Determine the time t when the ball reaches its second peak height. What is the height of the ball at this time? Between the second and third bounce 9. Find the function y = h3(t) for the height of the ball between the second and third bounce. 10. Find the time t3 when the ball hits the ground for the third time. The function h3(t) then describes the height of the ball over the time interval [t2, t3]. 11. Find the velocity v3(t) of the ball over the interval [t2, t3]. 12. Determine the time t when the ball reaches its third peak height. What is the height of the ball at this time? Graphs 13. On the coordinate systems below sketch accurate graphs of the height and velocity of the ball over the time interval [0, t3].

Answer 1

1.

Initially the ball is throw straight up with the velocity 78 ft/s from height 10 ft. So, by applying Newton's equation of motion

Lets up side is positive

So, equation of height with respect to floor at height of 10 ft is given as :

s(t) = ut + at²​​​​​​/2

Acceleration due to gravity is g=32 ft/s² downward so a= - g

s(t) =78t-(16)t²​​​​​​ ft

At t=0 it is at height of 10 ft so with respect to ground height h(t) is given as:

h(t) =s(t) + 10

h(t) =10 + 78t - 16t²​​​​​​

2.

The time when h(t) =0 is :

10 + 78t - 16t²=0

16t² - 78t - 10 = 0

t₁ = 5 sec

t₂ = - 0.125 sec

time can't be negative So

t₁= 5 sec

3.

Velocity between t=0 to t= =5 s is given as:

V(t) = U + at

Where, a= - g

V(t) =78 -32t

4.

When the ball reach at peak then V =0 so,

V(t) =78 - 32t =0

t= 2.4375 sec

h(2.4375) =10 + 78(2.4375) - 16(2.4375²​​​​​​)

h(2.4375) = 105.062 ft

5.

After the 1 collision velocity of ball becomes u=64 ft/s so height h₂(t) is given as:

h₂(t) = ut +at²/2

Where a=-g

h₂(t) = 64t - 16t²

Because 5 sec has been already taken before the first strike so h₂(t) become :

h₂(t) = 64(t-5) - 16 (t-5)²

6.

The ball will strike second time when h₂(t) =0

So,

64(t-5) - 16(t-5)² =0

Therefore, at t₂= 9 sec ball strike second time

7.

Initially velocity after first strike is u=64 ft/s

So,

V₂(t) = u + at

Where, a= - g

V₂(t) = 64 - 32t

Beacause ball has already taken 5 sec before first strike so

V₂(t) = 64 - 32(t-5)

8.

Ball will be at highest point when V(t) =0

Therefore,

V₂(t) =0= 64 - 32(t-5)

t=7 sec

So height of ball at t=7 sec is given as:

h₂(7) = 64(7-5) - 16 (7-5)²

h₂(7)=64 ft

9.

After third bounce the initial velocity of ball is u= 48 ft/s

So, Height h₃(t) is given as:

h₃(t) = ut +at²/2

Where a= - g

h₃(t) = 48t - 16t²

But we know that ball has already taken t₂= 9 sec before second strike

So height become:

h₃(t) = 48(t-9) - 16(t-9)²

10.

Ball will strike third time when

h₃(t) =0

So  

h₃(t) = 0 =48(t-9) - 16(t-9)²

So, t₃= 12 sec

11.

Initial velocity after second strike is u= 48 ft/s

So,

V₃(t) = u + at

Where a= - g

V₃(t) =48 - 32t

But ball has already taken t₂ =9 sec before second strike so,

V₃(t) = 48 - 32(t-9)

12.

The ball will be at highest point when V₃(t) = 0

So, V₃(t) = 0 = 48 - 32(t-9)

t= 10.5 sec

Therefore height at this time is :

h₃(10.5) = 48(10.5 - 9) - 16(10.5 - 9)²

h₃(10.5)= 36 ft

13.