Question

Someone please show me how to solve these two questions? I don't understand them.. Thank you..
Brittany drops a ball from a bridge at an initial height of 80 meters.
A. What is the height of the ball to the ball to the nearest tenth of a meter exactly 3 seconds after she releases the ball?
B. How many seconds after the ball is released will it hit the ground?

Answer 1

ZebraLover17 have you done solving of quadratics yet?

Answer 2

(delta)x = vt + 0.5at^2
delta x is how far it falls, v is starting velocity (0 m/s) and a is the acceleration (-9.8 m/s^2)
3 seconds after, t = 3
(delta)x = 3*0 + 0.5(-9.8)(3)^2
(delta)x = -44.1
so the ball is 80-44.1 meters above the ground.
for the second question
-80 = 0.5(-9.8)(t)^2
t=4.04 seconds

Answer 3

@ZebraLover17 if you check your material, I'd assume you'd have this equation there somewhere for this -> \(\bf \text{initial velocity formula}\\

\Large h = -4.9t^2+v_ot+h_o \qquad \text{in meters}\\
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object, in this case }80\\
h=\textit{height of the object at "t" seconds}\)

Answer 4

yeah. i looked in my book, but i dont know where to get v in the question.. I didnt understand any of it..

Answer 5

hmmm brittany is on a bridge and she's DROPPING THE BALL from the bridge
the ball is starting out with a velocity of 0 I'd think

Answer 6

so the equation should be: h(t)= 4.9t^2 + 0t + 80 ?

Answer 7

h(t)= -4.9t^2 + 0t + 80
^

Answer 8

Isn't that what I wrote?

Answer 9

though @jollyjolly0 used another formula, just as valid, and I wonder if better fit for this

Answer 10

ZebraLover17 the initial velocity uses a negative leading term, so it'd be -4.9

Answer 11

thank you

Answer 12

yw