BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.5, Problem 111E

(a)

To determine

The time taken by the ball to reach a height of 24ft if the ball is thrown straight upward at an initial speed of 40ft/s.

Expert Solution

Answer to Problem 111E

The ball reaches a height of 24ft after 1sand112s_.

Explanation of Solution

Given:

An object thrown or fired straight upward at an initial speed of v0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h=16t2+v0t.

Calculation:

The height of the ball is given as 24ft and the ball is thrown straight upward at an initial speed of 40ft/s.

Therefore, substitute 24 for h and 40ft/s for v0 in the equation h=16t2+v0t to obtain the time taken by the ball to reach a height of 24ft.

24=16t2+40t16t240t+24=0

Simplify further as follows.

8(2t25t+3)=02t25t+3=02t22t3t+3=02t(t1)3(t1)=0

On further simplifications, the following is obtained.

(t1)(2t3)=0t1=0or 2t3=0t=1or t=32t=1or t=112

Thus, the ball reaches a height of 24ft after 1sand112s_.

(b)

To determine

The time taken by the ball to reach a height of 48ft if the ball is thrown straight upward at an initial speed of 40ft/s.

Expert Solution

Answer to Problem 111E

The ball never_ reaches the height of 48ft.

Explanation of Solution

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form ax2+bx+c=0,a0 can be obtained by using the quadratic formula x=b±b24ac2a.

Calculation:

The height of the ball is given as 48ft and the ball is thrown straight upward at an initial speed of 40ft/s.

Therefore, substitute 48 for h and 40ft/s for v0 in the equation h=16t2+v0t to obtain the time taken by the ball to reach a height of 48ft.

48=16t2+40t16t240t+48=0

Simplify further as follows.

8(2t25t+6)=02t25t+6=0

Compare this equation with the general form ax2+bx+c=0.

Here a=2,b=5and c=6.

Use the quadratic formula to obtain the roots of the given equation.

t=(5)±(5)24×2×62×2t=5±25484t=5±234

Since the square of any real number is nonnegative, 23 is undefined in the real number system.

Thus, the ball never_ reaches the height of 48ft.

(c)

To determine

The greatest height reached by the ball.

Expert Solution

Answer to Problem 111E

The greatest height reached by the ball is 25ft_.

Explanation of Solution

Result used:

Discriminant of a quadratic equation:

The quantity b24ac is called the discriminant of a quadratic equation of the form ax2+bx+c=0,a0, from which the number of solutions the quadratic equation can have, can be obtained.

1. If b24ac>0, there are two unequal real solutions.

2. If b24ac=0, there is a repeated real solution, a double root.

3. If b24ac<0, there is no real solution.

Calculation:

Given that the height of an object thrown straight upward at an initial speed of v0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h=16t2+v0t.

Since the ball is thrown straight upward at an initial speed of 40ft/s, substitute v0=40ft/s.

16t240t+h=0

Compare this equation with the general form ax2+bx+c=0.

Here a=2,b=5and c=6.

Note that, the ball will reach the highest point only once and the quadratic equation has only one solution when the discriminant is zero.

Therefore, let the discriminant be zero.

D=b24ac0=(40)24×16×h0=160064h64h=1600

Simplify the above equation as follows.

h=160064h=25

Thus, the greatest height reached by the ball is 25ft_.

(d)

To determine

The time taken by the ball to reach the highest point.

Expert Solution

Answer to Problem 111E

The ball reaches the highest point after 114s_.

Explanation of Solution

The greatest height reached by the ball is obtained as 25ft and the ball is thrown straight upward at an initial speed of 40ft/s,

Substitute 25 for h and 40 for v0 in the equation h=16t2+v0t to obtain the time taken by ball to reach the highest point.

25=16t2+40t16t240t+25=0

Compare this equation with the general form ax2+bx+c=0.

Here a=16,b=40and c=25.

Use the quadratic formula to obtain the roots of the given equation.

t=(40)±(40)24×16×252×16t=40±1600160032t=4032

Simplify the above equation as follows.

t=54t=114

Thus, the ball reaches the highest point after 114s_.

(e)

To determine

The time taken by the ball to hit the ground.

Expert Solution

Answer to Problem 111E

The ball hits the ground after 212s_.

Explanation of Solution

The height of a ball is 0 when the ball hits the ground and the ball is thrown straight upward at an initial speed of 40ft/s.

Substitute 0 for h and 40ft/s for v0 in the equation h=16t2+v0t.

0=16t2+40t16t240t=0t(16t40)=0t=0or 16t40=0

Simplify further as follows.

t=0or 16t40=0t=0or 16t=40t=0or t=4016t=0or t=52

That is, t=0or t=212.

Thus, the ball hits the ground after 212s_.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!