The height s(t) and velocity v(t) at time t of an object tossed vertically in th

Question

The height s(t) and velocity v(t) at time t of an object tossed vertically in the air near the earth's surface (and ignoring all air resistance) is given by s(t)=s0+v0t-0.5gt^2 and v(t)=s'(t)=v0-gt, where s0=s(0) is the initial position at t=0, v0=v(0) is the initial velocity at t=0 and -g is the acceleration due to gravity. In this porblem we shall assume that s0=v0>0.

1) a.when does the object reach its maximum height?

b. what is the maximum heigh of the object?

c. when does the object hit the ground after being tossed?

d. what is the object's velocity when it hits the ground?

2) This problem extends the results in problem (1). Let c denote any positive number not exceeding the maximum height in (a). Intuituvely, there are two times when the height of the object is c. Namely at some time t1 when the object is rising and another time and another time t2 when the object is falling. (At the maximum height, we should have t1=t2.)

a. given c, show how to compute t1 and t2. Show where you use the assumption that c doesnt exceed the maximmum height of the object.

b. let v1=v(t1), the velocity of the object when it is at height c and rising. Let v2=v(t2), the velocity of the object when it is at height c and falling. Show v2=-v1.

Explanation / Answer

1a)The object reaches maximum height when v(t)=0

v0-gt1=0

t1=v0/g

b)maximum height=s(vo/g)=s0+v0^2/g-v0^2/2g=s0+v0^2/2g

c)t=2t1=2v0/g

d)v=-v0 (- sign indicates it is downward, v=v0-g*(2v0/g))

2a)c=s0+v0*t-0.5g*t^2

Solving for t,

t1= (v0-sqrt(-2 c g+2 g s0+v0^2))/g

t2= (sqrt(-2 c g+2 g s0+v0^2)+v0)/g

If c> hmax, the term inside the square root becomes negative and we get an imaginary time for that c since the ball does not pass through any point higher than the maximum height.

v(t1)=v0-gt1=v0-(v0-sqrt(-2 c g+2 g s0+v0^2))=sqrt(-2 c g+2 g s0+v0^2)

v(t2)=v0-gt2=v0-(sqrt(-2 c g+2 g s0+v0^2)+v0)=-sqrt(-2 c g+2 g s0+v0^2)

Hence -v(t1)=v(t2)

Hence proved

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