Answer
(a) After $1sec$ while thrown upwards and after $1.5sec$ while falling downwards.
(b) The ball never reaches $48feet$.
(c) The ball reaches its maximum height of $25feet$.
(d) The ball would take $1.25sec$ to reach $25feet$.
(e) It would hit the ground after $2.5sec$
Work Step by Step
We have a base formula: $h=-16t^2+v_{0}t$, but we are given initial speed of $v_{0}=40ft/s$, so we get a formula (which we will use later):
$h=-16t^2+40t$
(a) Now, we have to measure the time it takes for a ball to reach $24ft$ (which is value of $h$). But we also have to note that a ball reaches this height $2$ times, once when its thrown upwards and second time when it falls down to the ground.
$24=-16t^2+40t$
$-16t^2+40t-24=0$ //To simplify, we can divide the expression by -8
$2t^2-5t+3=0$
$t_{1} = \frac{5-\sqrt{1}}{4} =1 sec$
$t_{2}= \frac{5+\sqrt{1}}{4} = 1.5 sec$
(b) In this case $h=48$ :
$48=-16t^2+40t$
$16t^2-40t+48=0$ //To simplify, divide by 8
$2t^2-5t+6=0$
$t=\frac{5±\sqrt{-23}}{4}$, It is undefined here as we cannot take square root of a negative number, which means that the ball has never reached $48 feet$.
(c) Solution for a quadratic equation is always $2$ numbers, which means that the height reached in a specific amount of time gets the same number $2$ times (while thrown upwards and while falling downwards).
However, we get only one height value when the object reaches its maximum point. which is when a Discriminant ($D$) is $0$:
In case of: $-16t^2+40t-h=0$
$D=b^2-4ac$
$0=40^2-64h$
$64h=1600$
$h=25feet$
(d) As we found out in (c) maximum height is $25feet$, we will simply input this value to the quadratic formula we have and calculate time it takes to reach this height:
$25=-16t^2+40t$
$16t^2-40t+25=0$ //We already know from (c) that $D$=0, so we will instantly calculate $t$
$t=\frac{40}{32}=\frac{5}{4}=1.25sec$
(e) The ball hits the ground when:
$-16t^2+40t=0$
$t(40-16t)=0$
$t=0$ (Thrown point)
$40-16t=0$
$t=\frac{40}{16}=\frac{5}{2}=2.5sec$
