A stone is dropped from the roof of a building 256 ft high. Two seconds later a second stone is thrown downward
from the roof of the same building with an initial velocity of v 0 ft/s. If both stones hit the ground at the same
time, what is v 0 l
I For the first stone, j = 256 - I6t
2 . It hits the ground when 0 = s - 256 - 16?
2
, t
2 = 16, t = 4 seconds.
Since the second stone was thrown 2 seconds later than the first and hit the ground at the same time as the first, the
second stone's flight took 2 seconds. So, for the second stone, 0 = 256 + v a (2) - 16(2)
2
, v 0 = -192 ft/s.
17.32.
17.31 A woman standing on a bridge throws a stone straight up. Exactly 5 seconds later the stone passes the woman on
the way down, and 1 second after that it hits the water below. Find the initial velocity of the stone and the height
of the bridge above the water.
17.30
17.29
From Problem 17.27, we know that the object hits the ground u 0 /16 seconds after it was thrown. Hence,
With what velocity must an object be thrown straight up from the ground in order to reach a maximum height of h
feet?
With what velocity must an object be thrown straight up from the ground in order for it to hit the ground ? 0 seconds
later?
I By Problem 17.27, the object hits the ground after i> 0 /16 seconds. At that time, v = v 0 — 32? = v a —
32(i> 0 /16) = —D O . Thus, the velocity when it hits the ground is the negative of the initial velocity, and, therefore,
the speeds are the same.
17.28
Under the conditions of Problem 17.27, show that the object hits the ground with the same speed at which it was
initially thrown.
I s = s 0 + v 0 t - I6t
2 . In this case, s 0 = 0. So s = v 0 t - 16?
2
, v = D,s = v 0 -32t, a = D,v = D
2 s = -32.
So the unique critical number is t = i> 0 /32, and, since the second derivative is negative, this yields the
maximum height. Thus, the time of the upward flight is i> 0 /32. The object hits the ground again when
s = v 0 t — I6t
2 = 0, v 0 = 16?, t = u 0 /16. Hence, the total time of the flight was i> 0 /16, and half of that time,
i> 0 /32, was used up in the upward flight. Hence, the time taken on the way down was also D 0 /32.
17.27
f Since she is moving only under the influence of gravity, her height s = s 0 + v 0 t — 16?
2
. In this case, s a = 10
and va = 12. So 5 = 10 + 12? - 16?2, v = D,s = 12 - 32?, D2s = -32. (a) Setting v=Q, we obtain
t = 0.375. Since the second derivative is negative, this unique critical number yields an absolute maximum.
When ? = 0.375, s = 16.75 ft. (ft) To find when she hits the water, set s = 10+ 12? - 16r
2 =0. So (540(1 + 20 = 0, and, therefore, she hits the water at ? = 1.25 seconds, (c) At ? = 1.25, c=-28 ft/s.
A ball is thrown vertically upward. Its height s (in feet) after t seconds is given by s - 48? - 16?
2
. For which
values of ? will the height exceed 32 feet?
I We must have 48?-16?
2 >32, 3?-?
2 >2, ?
2 -3? + 2<0, (?-2)(?-1) <0. The latter inequality
holds precisely when 1< ? < 2.
The distance a locomotive is from a fixed point on a straight track at time ? is given by s = 3?
4 - 44?
3 + 144?
2
.
When was it in reverse?
I u = D,s = 12?
3 -132?
2 + 288?=12?(?
2 -ll? + 24) = 12?(?-3)(?-8). The locomotive goes backwards
when v<0. Clearly, v>0 when ?>8; u<0 when 3Q when 0 t < 0. Thus, it was in reverse when 3 < ? < 8 (and, if we allow negative time, when ? < 0).
An object is thrown straight up from the ground with an initial velocity v 0 ft/s. Show that the time taken on the
upward flight is equal to the time taken on the way down.
136
CHAPTER 17
17.25
17.26
I The height of the stone s = s a + v 0 t— I6t
2 , where s a is the height of the bridge above the water. When
t = 5, s = s 0 . So s a = s 0 + v 0 (5) - 16(5)
2
, 5i; 0 = 400, u 0 = 80ft/s. Hence, 5 = s 0 + SOt - 16t
2 . When
t = 6, s=0. So 0 = s 0 + 80(6) - 16(6)
2
, s 0 = 96ft.
f From Problem 17.27, we know that the object reaches its maximum height after i> 0 /32 seconds. When
t=v a /32, s = v a t-16t
2 = v
2
0 /64. Hence, h = v
2
0 /64, v 0 = 8VK.
f 0 = i; 0 /16, U 0 = 16V
from the roof of the same building with an initial velocity of v 0 ft/s. If both stones hit the ground at the same
time, what is v 0 l
I For the first stone, j = 256 - I6t
2 . It hits the ground when 0 = s - 256 - 16?
2
, t
2 = 16, t = 4 seconds.
Since the second stone was thrown 2 seconds later than the first and hit the ground at the same time as the first, the
second stone's flight took 2 seconds. So, for the second stone, 0 = 256 + v a (2) - 16(2)
2
, v 0 = -192 ft/s.
17.32.
17.31 A woman standing on a bridge throws a stone straight up. Exactly 5 seconds later the stone passes the woman on
the way down, and 1 second after that it hits the water below. Find the initial velocity of the stone and the height
of the bridge above the water.
17.30
17.29
From Problem 17.27, we know that the object hits the ground u 0 /16 seconds after it was thrown. Hence,
With what velocity must an object be thrown straight up from the ground in order to reach a maximum height of h
feet?
With what velocity must an object be thrown straight up from the ground in order for it to hit the ground ? 0 seconds
later?
I By Problem 17.27, the object hits the ground after i> 0 /16 seconds. At that time, v = v 0 — 32? = v a —
32(i> 0 /16) = —D O . Thus, the velocity when it hits the ground is the negative of the initial velocity, and, therefore,
the speeds are the same.
17.28
Under the conditions of Problem 17.27, show that the object hits the ground with the same speed at which it was
initially thrown.
I s = s 0 + v 0 t - I6t
2 . In this case, s 0 = 0. So s = v 0 t - 16?
2
, v = D,s = v 0 -32t, a = D,v = D
2 s = -32.
So the unique critical number is t = i> 0 /32, and, since the second derivative is negative, this yields the
maximum height. Thus, the time of the upward flight is i> 0 /32. The object hits the ground again when
s = v 0 t — I6t
2 = 0, v 0 = 16?, t = u 0 /16. Hence, the total time of the flight was i> 0 /16, and half of that time,
i> 0 /32, was used up in the upward flight. Hence, the time taken on the way down was also D 0 /32.
17.27
f Since she is moving only under the influence of gravity, her height s = s 0 + v 0 t — 16?
2
. In this case, s a = 10
and va = 12. So 5 = 10 + 12? - 16?2, v = D,s = 12 - 32?, D2s = -32. (a) Setting v=Q, we obtain
t = 0.375. Since the second derivative is negative, this unique critical number yields an absolute maximum.
When ? = 0.375, s = 16.75 ft. (ft) To find when she hits the water, set s = 10+ 12? - 16r
2 =0. So (540(1 + 20 = 0, and, therefore, she hits the water at ? = 1.25 seconds, (c) At ? = 1.25, c=-28 ft/s.
A ball is thrown vertically upward. Its height s (in feet) after t seconds is given by s - 48? - 16?
2
. For which
values of ? will the height exceed 32 feet?
I We must have 48?-16?
2 >32, 3?-?
2 >2, ?
2 -3? + 2<0, (?-2)(?-1) <0. The latter inequality
holds precisely when 1< ? < 2.
The distance a locomotive is from a fixed point on a straight track at time ? is given by s = 3?
4 - 44?
3 + 144?
2
.
When was it in reverse?
I u = D,s = 12?
3 -132?
2 + 288?=12?(?
2 -ll? + 24) = 12?(?-3)(?-8). The locomotive goes backwards
when v<0. Clearly, v>0 when ?>8; u<0 when 3Q when 0 t < 0. Thus, it was in reverse when 3 < ? < 8 (and, if we allow negative time, when ? < 0).
An object is thrown straight up from the ground with an initial velocity v 0 ft/s. Show that the time taken on the
upward flight is equal to the time taken on the way down.
136
CHAPTER 17
17.25
17.26
I The height of the stone s = s a + v 0 t— I6t
2 , where s a is the height of the bridge above the water. When
t = 5, s = s 0 . So s a = s 0 + v 0 (5) - 16(5)
2
, 5i; 0 = 400, u 0 = 80ft/s. Hence, 5 = s 0 + SOt - 16t
2 . When
t = 6, s=0. So 0 = s 0 + 80(6) - 16(6)
2
, s 0 = 96ft.
f From Problem 17.27, we know that the object reaches its maximum height after i> 0 /32 seconds. When
t=v a /32, s = v a t-16t
2 = v
2
0 /64. Hence, h = v
2
0 /64, v 0 = 8VK.
f 0 = i; 0 /16, U 0 = 16V