Question

A ball is thrown straight up at a speed of 2 m/s and is caught at some time later. Assume the ball is thrown and caught at the same height.
(c) What the the ball’s maximum height with respect to it’s initial position?
(d) How long does the ball take to get to it’s maximum height?
(e) What is the ball’s total flight time?
(f) What is the velocity of the ball when it is caught?
(g) Graph the position, velocity, and acceleration of the ball.

Answer 1

(a) At maximum height the vertical speed of any thrown object equals to zero.
(b) It's always -9.81m/s^2, because it's the acceleration constant on earth.
(c) find the time with the displacement formula (answers (d)) plug that time in the velocity formula
(e) use displacement formula to find how long it takes it to fall.
(f) = (d)+(e)

Answer 2

Need to calculate time to graph properly because all graphs are dependent on time.

Answer 3

this is the displacement formula right? s=ut+1/2at^2

Answer 4

What are s and u defined as?

Answer 5

\[y_f=y_i+v_it+\frac{ 1 }{ 2 } at^2\]
yf = final position
yi = initial position
vi = initial velocity

Answer 6

do we know what yf, yi, and vi is?

Answer 7

@Ankh

Answer 8

You'll have two unknowns in the displacement formula (yf and time) so it's better to use the velocity formula, since you know at maximum height the velocity is briefly zero.

Answer 9

\[v_f = v_i+at\]
\[0 = 2+(-9.8)t\]

Answer 10

yi and vi you know:
yi = 0
vi = 2m/s

Answer 11

for t i got 0.20