Question

Please help! I dont know how to do these :(

Answer 1

Hello There!!

Given

`h = - 16 {t}^(2) + 48t + 64`

Where:- h= Height of ball (given in feet[ft]) and,

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎t= time (in seconds [s])

To find:-

Time = (t)

Solution

We know,

`\text{When the ball hits the ground its height will be 0}`

`\therefore h(t) = 0`

`\implies - 16 t^(2) + 48t + 64 = 0 \\ \\ \fbox{taking - 16 as common multiple} \\ \\ \implies - 16(t {}^(2) - 3t - 4) = 0 \\ \\ \implies t {}^(2) - 3t - 4 = 0 \\ \\ \fbox {factorising the \: equation} \\ \\ \implies t {}^(2) - t - 4t - 4 = 0 \\ \\ \implies t(t + 1) - 4(t + 1) = 0 \\ \\ \implies(t - 4)(t + 1) = 0`

`\tt{Either} \\ t - 4 = 0 \\ t = \red{ 4 } \\ \\ \tt{Or} \\ t + 1 = 0 \\ t = \red{ - 1}`

We know, Time cannot be negative.

So, Time (t) = 4 seconds

Answer 2

t = 4 s

Explanation:

Given function:

`h(t)=-16t^2+48t+64`

where:

• h = height of the ball (in feet)
• t = time (in seconds)

When the ball hits the ground, its height will be 0 ft.

Therefore, set the function to zero and solve for t:

`\begin{aligned}h(t) &=0\\ \implies -16t^2+48t+64 & =0\\ -16(t^2-3t-4)& =0\\ t^2-3t-4 &=0\\ t^2+t-4t-4&=0\\ t(t+1)-4(t+1)&=0\\(t-4)(t+1)&=0\\ \implies t&=4, -1\end{aligned}`

As time is positive, t = 4 s (only).