A projectile fired upwards on earth and mars.

[m0286]

Hello
I have a calculus problem (1st year) that I cannot solve I am completely confused:
It says:
We are given that the Earth's gravitational acceleration is gE=9.8m/s^2 (downward) and that on Mars gM=3.72m's^2 (downward)
If a projectile is fired upwards from the Earth it lands back on Earth in 10seconds. How long will the projectile take to land back on the surface of Mars if it is fired upwards with the same initial velocity?

So what I've done is:
Earth: d(t)=1/2at^2 so x(t)=490m... so would this be the entire distance or just the distance downward, so that the entire distance would be 980?

Then since velocity =d/t its 980/10s =98m/s (or) 490/10 =49m/s... (not sure which one to use)
And d(t)=vt+1/2at^2 So would I just take the values I got.. and plug them and the Mars acceleration in this equation and solve for t?? Or I have I done something completely wrong? Thanks

 

[HallsofIvy]

Hello
I have a calculus problem (1st year) that I cannot solve I am completely confused:
It says:
We are given that the Earth's gravitational acceleration is gE=9.8m/s^2 (downward) and that on Mars gM=3.72m's^2 (downward)
If a projectile is fired upwards from the Earth it lands back on Earth in 10seconds. How long will the projectile take to land back on the surface of Mars if it is fired upwards with the same initial velocity?

So what I've done is:
Earth: d(t)=1/2at^2 so x(t)=490m...

Okay, that's wrong! That would be for something that had initial velocity 0. It would fall a distance (1/2)at^2 in time t. The whole point here is that the object has an initial velocity, upward, otherwise it wouldn't move upward at all. Of course, you aren't told that initial velocity- that's why you are told that the time it takes to come back to earth- so you can calculate that.

Since this was posted in the "Calculus and beyond" section, I presume that you know how to do this:
dv/dt= a so v= dx/dt= at+ v₀, where v₀ is the initial velocity. Integrating again, x(t)= (1/2)at²+ v₀t+ x₀. Taking the initial height, x₀, to be 0, we have just x(t)= (1/2)at²+ v₀t. Obviously, when t= 0, x(0)= 0. But you are also told that x(10)= 0. What must v₀ be for that to be true (with a= -980 of course)?

Now replace a with -3.72, use that same v₀ and solve (1/2)at²+ v₀t= 0 for t. (That has one obvious answer and the other is easy.)

so would this be the entire distance or just the distance downward, so that the entire distance would be 980?

Then since velocity =d/t its 980/10s =98m/s (or) 490/10 =49m/s... (not sure which one to use)
And d(t)=vt+1/2at^2 So would I just take the values I got.. and plug them and the Mars acceleration in this equation and solve for t?? Or I have I done something completely wrong? Thanks

 

[m0286]

Thanks!

ohhhh
Makes so much more sense to me now, i got it.. Thanks a lot!