13. Answer: The height will be maximum 372.25 ft at 4.63 s.
Explanation:
Given that,
Upward velocity v = 148 ft/s
Function
\(h = -16t^2+148t+30\).....(I)
On differentiate
\(\dfrac{dh}{dt}=-32t+148\)....(II)
for maximum height,
\(\dfrac{dh}{dt}=0\)
Put the value of \(\dfrac{dh}{dt}\) in equation (II)
\(-32t+148=0\)
\(t = \dfrac{148}{32}\)
\(t = 4.63\ s\)
The maximum height is
Put the value of t in equation(I)
\(h = -16\times(4.63)^2+148\times4.63+30\)
\(h = 372.25 ft\)
Hence, The height will be maximum 372.25 ft at 4.63 s.
16. Answer: The value of x is 2, \(\dfrac{1}{2}\).
Explanation:
Given that,
\((2x-4)(2x-1)=0\)
Using zero product property
\(2 x-4=0\)
\(x = 2\)
And,
\(2x-1=0\)
\(x = \dfrac{1}{2}\)
Hence, The value of x is 2, \(\dfrac{1}{2}\).
