Question

A ball is thrown directly downward with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does it strike the ground?

Answer 1

Initial velocity (Vi) = 8m/s
Acceleration due to gravity (a) = 9.8 m/s^2
Distance (d)= 30m

First calculate the velocity at the time of impact:
\[V_f^2=V_i^2 + 2ad=(8m/s)^2 + 2(9.8m/s^2)(30m)=652 m/s\]\[V_f=\sqrt{652m/s}=25.5m/s\]
Now that you know the final velocity, you can calculate the time using another kinematic equation:
\[d=\frac{V_i+V_f}{2}t\]\[30=\frac{(8m/s + 25.5m/s)}{2}t\]\[t=1.79s\]

Answer 2

you can just use the fomula \[\huge x - x_0 = V_0 t + \frac 12 at^2\]
where:
x = final height
x _ 0 = initial height
V_ 0 = initial speed
t= time
a = acceleration

in this case x would be 30 and x_0 would be 0 because obvioulsy you START at the origin so you start with 0 height and then it says the height from where the ball was thrown is 30

then, the v_0 would be 8 m/s because it says the initial speed is 8 m/s

and the acceleration would be 9.8 m/s^2 because it is the acceleration due to gravity

so substitute
\[\huge 30 - 0 = (8)t + \frac 12 (9.8)t^2\]
simplify that..
\[\huge \implies 30 = 8t + 4.9t^2\]
subtract 30 from both sides..
\[\huge \implies 4.9t^2 + 8 t - 30 = 0\]
use quadratic formula to get the value of t
\[\LARGE t = \frac{-8 \pm \sqrt{(8)^2 - 4(4.9)(-30)}}{2(4.9)}\]
\[\huge \implies t \approx 1.78921 \implies t \approx 1.79 s\]
does that help?

Answer 3

Pfft....I wanted to avoid solving a quadratic :P

Answer 4

wimp :P lol

Answer 5

Yes...yes I am !

Answer 6

Yes the above solutions give the same answer so either of those would do.