Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

please someone kindly and neatly help me out with this one please thank you ston

ID: 2885609 • Letter: P

Question

please someone kindly and neatly help me out with this one please thank you

stone thrown vertically Suppose a stone is thrown vertically up- ward from the edge of a cliff with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height s (in ft) of the stone above the ground t seconds after it is thrown is s1612 64t + 3.2. a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground? f. On what intervals is the speed increasing? 17. A

Explanation / Answer

(a)

Determine the velocity v of the stone after t seconds.

SOLUTION: v(t) = h 0 (t) = ?32t + 64

(b)

When does the stone reach its highest point?

SOLUTION: The highest point is when the velocity is zero.

Set the velocity equal to zero and solve for t.

0 = ?32t + 64 so t = 2

To find the height at time t = 2,

(c)

what is its height ?

evaluate h(2) = ?16(2)2 + 64(2) + 32 = 96 ft.  

(d)

When does the stone strike the ground?

SOLUTION:

To find the point at which the stone strikes the ground,

set the height equal to zero and solve for t.

0 = ?16t 2 + 64t + 32 ? t = 2 ± ? 6 Because 2 ? ? 6 < 0,

we can ignore that solution and only use t = 2 + ? 6 seconds.

(e) what is its velocity at that point

To find the velocity, we evaluate

v(2 + ? 6. v(2 + ? 6) = ?32(2 + ? 6) + 32 = ?32? 6ft/sec.

(f)

On what intervals is its speed increasing?

SOLUTION:

It stands to reason that the speed is only increasing as the rock is falling, on the interval (2, 2 + ? 6).

We can determine that analytically by finding the derivative of the speed, and see when its derivative is negative. speed(t) = |v(t)| = {?32t + 64 if 0 < t ? 2

= {32t ? 64 if 2 < t < 2 + ? 6

And the first derivative is defined piecewise as well

speed'(t) = { ?32 if 0 < t < 2

= { 32 if 2 < t < 2 + ? 6

So we have that the speed is increasing on the interval (2, 2 + ? 6), as we expected