A) a dynamic blast blows a heavy rock straight up with a launch velocity of 160 ft/sec(about 109mph). It reaches a height of = 160 16 after . (i) how high does the rock go? (ii) what are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down? (iii) what is the acceleration of the rock at any time during its flight (after the blast)? (iv) when does the rock hit the ground again?
Answers
v(t) = s' (t) = -32 t + 160
a(t) = v' (t) = -32
a)
Setting v(t) = 0
-32 t + 160 = 0
-32 t = -160
t = 5 seconds
Height: s(5) = -16(5)^2 + 160(5)
Height = 400 ft <==== Answer to A
b)
256 = -16(t)^2 + 160t
Divide by 16
16 = -t^2 + 10 t
t^2 - 10t + 16 = 0
Factor
(t-8)(t-2)
At t= 2 and at t= 8, height is at 256 ft
V(2) = -32 t + 160
V(2) = 96 ft/second (upward)
V(8) = -32 t + 160
V(8) = -96 m/s (or 96 m/s downward)
C) acceleration:
a(t) = v'(t) = s"(t) = -32
a(t) = -32 ft/sec^2
(i) The maximum height the rock goes is 400 ft.
(ii) The velocity of the rock when it is 256 ft above the ground on the way up is 96 ft/s
velocity of the rock when it is 256 ft above the ground on the way down is - 96 ft/s
(iii) The acceleration of the rock at any time during its flight (after the blast) is 32 ft/s².
(iv) The rock hits the ground again after 10 s.
Given: The launch velocity is 160 ft/sec.
It reaches a height of S = 160 t - 16 t².
To Find:
(i) the maximum height the rock goes?
(ii) what are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
(iii) what is the acceleration of the rock at any time during its flight (after the blast)?
(iv) when does the rock hit the ground again?
Solution:
- We know that at the maximum height, the final velocity is zero.
(i) S = 160 t - 16 t² ....... (1)
⇒ v = dS/dt = 160 - 32 t ....... (2)
Now v = 0 for final velocity, so;
⇒ 160 - 32 t = 0
⇒ t = 5 sec
So, at t = 5 s, it reaches the maximum height. To find the maximum height, we shall substitute t = 5 s in S = 160 t - 16 t².
So, S max = 160 × 5 - 16 × ( 5 )²
= 400 ft.
Hence, the maximum height the rock goes to is 400 ft.
(ii) We are required to find the velocity of the rock when it is at a height of 256 m. So, we shall put the value in the relation, S = 160 t - 16 t².
256 = 160 t - 16 t²
⇒ t² - 10t + 16 = 0
⇒ ( t - 2 ) ( t - 8 ) = 0
⇒ t = 2, 8.
Hence, we can see that at t = 2 s and t = 8 s, the rock is 256 m above the ground.
Putting the values of t in (2), we get;
v (2) = 160 - 2 × 32 = 96 ft/s
v (2) = 160 - 8 × 32 = - 96 ft/s
Hence, the velocity of the rock when it is 256 ft above the ground on the way up is 96 ft/s
velocity of the rock when it is 256 ft above the ground on the way down is - 96 ft/s
(iii) We need to find the acceleration equation using differentiation,
a (t) = - 32 m/s² [ negative sign is due to deceleration ]
Since the equation is independent of the value of t, so the acceleration of the rock at any time during its flight is 32 ft/s².
(iv) When the rock hits the ground again, the initial velocity of the rock must be zero. Putting the value of velocity to zero in (2), we get;
160 - 32 t = 0
⇒ t = 160/32 s
= 5 s
Now, this is the time taken by the rock to reach the ground from the highest point, so the time taken by the rock to reach the highest point is also 5 s.
So, total time = 5 s + 5 s
= 10 s
Hence, the rock hit the ground again after 10 s.
Compiling all the answers we get,
(i) The maximum height the rock goes is 400 ft.
(ii) The velocity of the rock when it is 256 ft above the ground on the way up is 96 ft/s
The velocity of the rock when it is 256 ft above the ground on the way down is - 96 ft/s
(iii) The acceleration of the rock at any time during its flight (after the blast) is 32 ft/s².
(iv) The rock hits the ground again after 10 s.
#SPJ3