Question

A toy cannon ball is launched from a cannon on top of a platform. The equation h(t) = -5t^2 + 20t + 4 gives the height h, in meters, of the ball t seconds after it is launched.

A. What equation can be used to tell whether the ball reaches a height of 12 m ?
B. Does the ball reach a height of 12m? (yes or no)

Answer 1

Answer:

A) $5t^{2} -20t+8$  =0

B)t= 0.45 seconds or t= 3.5 seconds

If time is greater than 0.45 seconds then ball will reach height 12m and higher.

Step-by-step explanation:

Given Equation:

h(t)=$-5t^{2} +20t+4$ ------------------------------(Equation 1)

a) Equation to tell if ball reaches height of 12m . for that :

h= 12m

put in Equation 1.

12 = $-5t^{2} +20t+4$

or $-5t^{2} +20t+4$ =12

or $-5t^{2} +20t+4$ -12 = 0

$-5t^{2} +20t-8$=0

or $5t^{2} -20t+8$  = 0 -----------------------------------(Equation 2)

This Equation tells if ball reaches height of 12m

b) Does ball reaches height of 12m :

For that, the value of time can be found out from the equation above,

$5t^{2} -20t+8$  = 0

It can be solved using the quadratic formula:

$t= \frac{(-b) +_{-} \sqrt{b^{2} -4ac }  }{2a}$

$t= \frac{(20) +_{-} \sqrt{(-20)^{2} -4(5)(8) }  }{2(5)}$

t= 0.45 seconds or t= 3.5 seconds

If time is greater than 0.45 seconds then ball will reach height 12m and higher.

Answer 2

The maximum height reached is greater than 12m, hence the ball reach a height of 12m

The height of the toy cannon all launched is expressed as $h(t) = -5t^2 + 20t + 4$

h is the height reached by the ball

t is the time taken by the ball to reach its maximum height

A) In order to determine whether the ball reached the height of 12m, we will have to substitute h = 12 into the formula as shown:

$12= -5t^2 + 20t + 4\\ -5t^2 + 20t + 4 -12=0\\ 5t^2 - 20t +8=0$

Hence the equation that can be used to tell whether the ball reaches a height of 12 is $5t^2-20t+8=0$

B) At the maximum height

dh/dt = 0

-10t + 20 = 0

-10t = -20

t = 20/10

t = 2

Get the maximum height reached

h(2) = -5(2)^2 + 20t + 4

h(2) = -5(4) + 20t + 4

h(2) = -20 + 20(2) + 4

h(2) = -20+40+4

h(2) = 24 m

Since the maximum height reached is greater than 12m, hence the ball reach a height of 12m

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