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

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

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.

thank you Bob!

had a resembling question :)

To solve the given questions, let's break them down step-by-step:

a) What is the initial velocity of the ball?

In this case, we can analyze the motion of the ball using the equation:

Y - Yo = Vyo(t) + 0.5G(t)^2

where:
Y is the final height (0 m in this case, as the ball returns to its starting point)
Yo is the initial height (also 0 m)
Vyo is the initial velocity in the y-direction
G is the acceleration due to gravity (9.8 m/s^2 on Earth)
t is the time taken (5 seconds, as given in the question)

Rearranging the equation, we have:

Vyo = (Y - Yo - 0.5G(t)^2) / t

Plugging in the values:
Vyo = (0 - 0 - 0.5 * 9.8 * (5)^2) / 5
Vyo = 24.5 m/s

So the initial velocity of the ball is 24.5 m/s.

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

To find the maximum height (Ymax), we need to find when the vertical velocity (Vy) becomes zero. Using the equation:

V^2 = Vo^2 + 2G(y-yo)

where:
V is the final velocity (0 m/s, as the ball reaches its highest point)
Vo is the initial velocity (24.5 m/s, as found in question a)
G is the acceleration due to gravity (9.8 m/s^2)
y is the final height (we need to find this)
yo is the initial height (0 m)

Rearranging the equation, we have:

y = (V^2 - Vo^2) / (2G) + yo

Plugging in the values:
y = (0 - (24.5)^2) / (2 * 9.8) + 0
y = -600.25 / -19.6
y = 30.625 m

Therefore, the maximum height reached by the ball is 30.625 meters.

c) If the ball is thrown on the moon with a different acceleration due to gravity (g_moon = 1.63 m/s^2) but the same initial velocity (24.5 m/s), what would be the air time and the max height reached by the ball?

To solve this, we can use the same equations as in parts a) and b), but with the new acceleration due to gravity.

a) Air time (t):
Using the equation from part a):
t = (Y - Yo - 0.5G(t)^2) / Vyo
t = (0 - 0 - 0.5 * 1.63 * (5)^2) / 24.5
t = -10.25 / 24.5
t = -0.418 s (approx.)

b) Maximum height (Ymax):
Using the equation from part b):
Ymax = (V^2 - Vo^2) / (2G) + yo
Ymax = (0 - (24.5)^2) / (2 * 1.63) + 0
Ymax = -600.25 / 3.26
Ymax = -184.11 m (approx.)

Therefore, on the moon with an acceleration due to gravity of 1.63 m/s^2, the air time is approximately -0.418 seconds (which doesn't make physical sense) and the maximum height reached by the ball is approximately -184.11 meters (which also doesn't make physical sense). It is important to note that this discrepancy arises due to the different acceleration due to gravity on the moon.