# 1d dynamics

A ball is thrown with an initial speed of v0=39.2m/s vertically to Mars. Its initial height is h=15m from the surface of mars. the acceleration of gravity on Mars is g=3.7m/s^2

(1) what is the maximum height hmax of the ball

(2) How long is the ball in the 'atmosphere'

(3) What is the speed vG of the ball when it hits the surface of Mars

(4) What is the height of the ball after exactly 5s?

(5) When the ball's height is 7m, what is the speed?

This is my homework. Could anyone explain this in the EASIEST way possible? :(

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Rocket50
Gold Member
Can you show what you have tried so far?

@Rocket50

Sure, this is my very first baby thread and I almost have no background in Physics so please don't make fun of me haha

(1) So, at maximum height, v=0m/s
First I'm tried to find t
x=1/2at+vt
v=at+v0
=gt+32.9=0
gt=-32.9
t=-32.9/g
t=-3.4

it takes -
and then i don't know what to do from here

honestly thats all i hace so far.. :( i can try to work on it more and come back
we can try to do this one question by one question.. :)

thanks

(1)

yf=y0+voyt+1/2at^2?

@CWatters Hey thanks for the reply! No i am in PHYS101 so no not yet. :(

Or is this an elementary physics equation lol

I googled questions asking for maximum height and I think this person used the SUVAT equation?

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462chevelle
Gold Member
Make sure when you use these equations, you use the velocity in context with respect to its meaning. Avg, initial, final, inst.

(1) So, at maximum height, v=0m/s
First I'm tried to find t
x=1/2at+vt
v=at+v0
=gt+32.9=0
gt=-32.9
t=-32.9/g
t=-3.4

If upwards velocity is a positive number, then acceleration downwards is a negative number.
t = 3.4 s
correct formula where to plug that time: x =1/2at²+vt

Then I guess one might calculate various things regarding a ball falling down from height x.

CWatters
Homework Helper
Gold Member
A ball is thrown with an initial speed of v0=39.2m/s vertically to Mars. Its initial height is h=15m from the surface of mars. the acceleration of gravity on Mars is g=3.7m/s^2
(1) what is the maximum height hmax of the ball
For this one I would use

V^2 = U^2 +2as

which doesn't involve time at all.