Question

This can either tank my grade or make it fly through the roof, so please try to be accurate with the answer :)) ​

Answer 1

Answer:

c) h = -16(t - 2)² + 74

Step-by-step explanation:

We're asked to represent height reached by the ball as a function of time.

The question also informs us about how the relationship will be a quadratic equation.

Mostly, quadratic equations are of the form

y = ax² + bx + c

So let the required relation be

h = at² + bt + c . . . . . . (¡)

where,

h is the height, t is the time and a, b, c are three different constants that are required to form the relationship.

We need three quantities so we'll require three different equations.

#1

The trajectory starts with ball being at a height of 10 ft. when at rest.

=> h = 10ft. when t = 0

fitting this in eqn. (¡) :-

=> h = at² + bt + c

=> 10 = a(0) + b(0) + c

=> 10 = c

h = at² + bt + 10 . . . .  .. (¡¡)

#2

The ball is at a height of 74 ft. when the time has elapsed to 2 seconds.

=> h = 74 ft. at t = 2s

substituting these values in eqn. (¡¡)

=> h = at² + bt + 10

=> 74 = a(2)² + b(2) + 10

=> 74 - 10 = 4a + 2b

=> 64 = 2(2a + b)

=> 32 = 2a + b

=> 32 - 2a = b . . . . . . (¡¡¡)

Hence, we obtained a relationship between a and b.

#3

It says the max height reached by the ball is 74 ft. and that too exactly after the ball was hit,

=> hₘₐₓ = 74 ft. at time 2s

You must know that in order to find maximum value of a function, we differentiate that function with respect to its variable and equate it to zero.

(why?)

'cause the slope(of the tangent to the curve) ,i.e., the derivative, at maxima or minima is zero.

=> d/ dt (h) = d/ dt (at² + bt + 10) = 0

=> dh/ dt = 2at + b = 0

=> 2at + b = 0

=> 2a(2) + b = 0

=> 2a + b = 0

=> b = -4a

equating this value of b with that obtained in eqn. (¡¡¡)

=>  32 - 2a = -4a

=> 4a - 2a = -32

=> 2a = - 32

=> a = -16

substituting the value of a thus obtained in eqn. (¡¡¡)

=> b = 32 - 2(-16)

=> b = 32 + 32

=> b = 64

Final equation:-

h = - 16t² + 64 + 10

Oh sad! we have a problem here. The options given in the question are in the form of perfect squares. So, solving it a bit more.

• taking -16 common from the first two variables

=> h = -16(t² + 4) + 10

• adding and subtracting 4 inside the braces so as to form the perfect square of (t + 2)

=> h = -16($\mathrm{\overline{t^2+4t+4}}$ - 4) + 10

• since, a² + b² + 2ab = (a + b)²
• taking -4 outside of the braces and multiplying it with -16 in the process

=> h = -16(t + 2)² -4(-16) + 10

=> h = -16 (t + 2)² + 64 + 10

=> h = - 16(t + 2)² + 74

That is option C.

                                         

DERIVATIVES USED :-

d/ dx (x²) = 2x

d/ dx (px) = p (p is some constant)

d/ dx (constant) = 0

                                           

This is what its trajectory will look like.