Question

1. A dynamite blast blows a heavy rock straight up with a launch velocity of 140 ft sec (about 109
mph). It reaches a height of <=101-12 1 after I sec.
How high does the rock go?
(b) What are the velocity and speed of the rock when it is 260 ft above the ground on the way
On the way down?
© What is the acceleration of the rock at any time t during its flight after the blast 2
d) When does the rock hit the ground again? ​

Answer 1

Answer

AnswerWe know, v=

AnswerWe know, v= dt

AnswerWe know, v= dtds

AnswerWe know, v= dtds

AnswerWe know, v= dtds =160−32t.

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96v(8)=160−256=−96

AnswerWe know, v= dtds =160−32t.We now find values of t for which s(t)=256 So160t−16t 2 =256⇒16(t 2 −10t+16)=0⇒(t−2)(t−8)=0⇒t=2, t=8.So v(2)=160−32.2=96v(8)=160−256=−96So the velocity on the way up in 96 m/s.