Question

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=32+64t−16t2.
What is the maximum height the ball reaches?
What is the velocity of the ball when it hits the ground (height 0)?

Answer 1

Answer:

a) 96ft

b) -78.4ft/s

Explanation:

Given the height of the ball modelled as s(t)=32+64t−16t²

a) at maximum height, the velocity of the ball is zero

Velocity = d(s(t))/dt = -32t+64

Since v = 0m/s at maximum height

0 = -32t+64

32t = 64

t = 64/32

t = 2seconds

This means it takes 2 seconds for the ball to reach its maximum height.

The maximum height reached will occur at t = 2secs

s(2) =32+64(2)-16(2)²

s(2) = 32+128-64

s(2) = 96ft

b) When the ball hits the ground the height = 0

s(t) = 0

-16t²+64t+32 = 0

t²-4t-2 = 0

Factorizing the expression:

-(-4)±√(-4)²-4(1)(-2)/2(1)

= 4±√16+8/2

= 4±√24/2

= (4+2√6)/2 and (4-2√6)/2

= 2+√6 and 2-√6

t= 4.45s and -0.45s

Substitute t = 4.45s into the equation for calculating the velocity

V(t) = -32t+64

V(4.45) = -32(4.45)+64

V(4.45) = -142.4+64

V(4.45) = -78.4ft/s

Answer 2

Answer:

a) $s_{max} = 96\,ft$, b) $v(4.449\,s) = -78.368\,\frac{ft}{s}$

Explanation:

a) The maximum height is obtained with the help of the First and Second Derivative Tests:

First Derivative

$v(t) = 64 - 32\cdot t$

$64 - 32\cdot t = 0$

$t = 2\,s$

Second Derivative

$a(t) = -32$ (absolute maximum)

The maximum height reached by the ball is:

$s (2\,s) = 32 + 64\cdot (2\,s) - 16\cdot (2\,s)^{2}$

$s_{max} = 96\,ft$

b) The time required by the ball to hit the ground is:

$32+64\cdot t - 16\cdot t^{2} = 0$

$-16\cdot (t^{2}-4\cdot t - 2) = 0$

$t^{2}-4\cdot t - 2 = 0$

$(t -4.449)\cdot (t+0.449)\approx 0$

Just one root offers a solution that is physically reasonable:

$t = 4.449\,s$

The velocity of the ball when it hits the ground is:

$v(4.449\,s) = 64 - 32\cdot (4.449\,s)$

$v(4.449\,s) = -78.368\,\frac{ft}{s}$