Question

I just want to be sure if my answers make any sense!

A kid throws a ball in the air and catches it 5 seconds later

a) What is the initial velocity of the ball? No air friction.

b) What is the maximum height which the ball has reached above its departure point?

c) If this kid throws the ball on the moon (g_moon = 1.63 m/s^2) and if he threw it at the same velocity as found in a), what would be the air time and what would be the max height reached by the ball?

a) Yo = 0 m
Y = 0 m
Vyo = ?
Vy = 0 m/s
G = 9.8 m/s^2
t = 5 s.

Y - Yo = Vyo(t) + 0.5G(t)^2
Vyo(t) = y - yo - 0.5G(t)^2
Vyo = y - yo - 0.5G(t)^2 / t
Vyo = 0 - 0 - 24.5
Vyo = 24.5 m/s

b) Y = Ymax when Vy = 0

V^2 = Vo^2 + 2G(y-yo)
y = (v^2-Vo^2 / -2G) + yo
Ymax = (0 m/s - 24.5 m/s)^2/ -2(9.8 m/s^2)
+ 0
Ymax = -600.25 m/s / -19.6 m/s^2
Ymax = 30.625

c) same thing as in a) and b) just a different g? Or is the more to it?

THank you

Answer 1

*at which the ball reached, sorry for my bad grammar

Answer 2

a. another way to check:
vf=g*t assume it is falling from the top.
Vf=9.8*2.5 Your answer looks good.

b. find the distance fell in 2.5sec
d=1/2 g t^2=4.9*6.25 looks good

c. just a different g.

Answer 3

thank you Bob!

Answer 4

had a resembling question :)