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

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071 ### 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 24 ft if the ball is thrown straight upward at an initial speed of 40ft/s.

Expert Solution

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 48 ft if the ball is thrown straight upward at an initial speed of 40ft/s.

Expert Solution

The ball never_ reaches the height of 48ft.

### Explanation of Solution

Formula used:

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

The greatest height reached by the ball is 25ft_.

### Explanation of Solution

Result used:

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

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

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_.

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