Calculus Question, who knows, come in!?

Question

The height of an objec above the ground at time t is given by

s=V0t - ((g/2) x t squre)

where V0 is the intial velocity and g is the accelereation

a) At what height is the object initially?

b) How long is the object in the air before it hits the ground?

c) When will the object reach its maximum height?

d) What is that maximum height?

Answer 1

a. put t =0 in your expression and you will get the answer

b. set your expression to 0 and solve for t. you will get two values for t. One is zero, the other is your answer

c. set the derivative of your expression and set it to zero. now solve for t.  whatever you get is the answer

d. plug in the t value you found in part c into your original expression and you will arrive at your answer

Answer 2

s = v0t - ½ gt²

a)

The height of the object initial is where it is when you sub t = 0 into the equation

Which in this case = 0

b) find how long until the object is at 0 again

s = v0t - ½ gt²

s = t(v0 - ½ gt)

it hits 0 again when v0 - ½ gt = 0

v0 = ½ gt

t = 2v0/g

c)

Max height is when derivative = 0

s ' = v0 - gt = 0

gt = v0

t = v0/g

d)

Max height is when you sub time of max height into the displacement equation

s = v0t - ½ gt²

s = v0(v0/g) - ½ g(v0/g)²

s = v0²/g - ½v0²/g

s = ½v0²/g

Answer 3

(a) initially at zero? plug in t=o, we get s=0

(b) Find out when does s=0(at ground) by factoring

factor out a t, we get

t * (Vo - g/2 * t) = 0

t=0 Or when t = 2Vo / g

(c) maximum height will occur half way between the time. It takes the same amount of time to go up as down.

when t = Vo / g

(d) plug in that value of t into the original s equation:

s=V0t - ((g/2) x t squre)

s = Vo(Vo/g) - (g/2) (Vo/g)squared

s = Vosquared / g - Vosquared / (2g)

s = (Vo squared) over (2g)

Answer 4

So we are given the height of an object above the ground at time t.  This is given by the function:

s(t) = V0(t) - (1/2)gt^2

In actuality, the position function of an object is usually defined as:

s(t) = V0(t) - (1/2)gt^2 + s0, where s0 is the initial height.  since that term is missing here, we know that the answer to question a) is that the intitial height is zero.

Another way to see that is to replace t=0 since we are at initial condition.

b)we want to know the time it takes to reach the ground again.  So we can set s = 0 (position above ground would be zero when it hits ground)

0 = v0(t) - (1/2)gt^2  factor out a t

0 = t(v0 - 1/2gt)

set each term to zero

t = 0  OR  v0 - 1/2gt = 0

the first one, t = 0, can't be our answer because that refers to the intital time.  we solve the 2nd equation

v0 - 1/2gt = 0

v0 = 1/2gt

t = 2v0/g

c)when it reaches maximum height, we know that the velocity will be zero.  take derivative (derivative of position gives us velocity) and set to zero and solve for t

s'(t) = v(t) = v0 - gt

v0 - gt = 0

v0 = gt

t = v0/g

notice that this time is half the time it takes to hit the ground.  This should make sense since projectile motion is parabolic and therefore symmetric.

d)max height can be found by taking the time in answer c) and plugging into position function:

s_max = v0(v0/g) - (1/2)g(v0/g)^2

= v0^2/g - (1/2)g(v0^2/g^2)

= (v0^2/g) - (1/2)(v0^2/g)

= (1/2)(v0^2/g)

voila