How Do You Calculate the Maximum Height of a Projectile Shot Upwards?

[pureouchies4717]

A projectile is shot straight up from the Earth's surface at a speed of 6000 km/hr.How high does it go?

i started out by converting km/hr to m/s

so i got 6000km/hr= 1666.666666m/s

then i used this formula:

v^2=2gy

and i got:
y= 141723 m= 141.723km (i just got lucky that my answer was close)

the correct answer wass 145 km

how do you get 145? can someone please help? thanks!

 

[Da-Force]

A projectile is shot straight up from the Earth's surface at a speed of 6000 km/hr.How high does it go?

6000km/hr= 1666.666666m/s

So.. there's only 1 direction for velocity, the y-direction.

$$y = V_i t + \frac{1/2} a_g t^2$$
and
$$V_f = V_i + a_g t$$

So it's a simple kinematics problem with a kick in it if I'm doing it correctly..

So I proceed to find t=170 seconds.. Plug it in and I get 141780 meters or 141.780 km... My god

My problem is that if we were talking about gravitational acceleration past the Earth, we would be given the height but... Well.. yeah

It might be that they rounded... *shrugs* See if anyone else gets a better idea.

 

[pureouchies4717]

can anyone else help please? I am positive that i did it wrong and just got lucky

 

[nrqed]

can anyone else help please? I am positive that i did it wrong and just got lucky

Have you covered $F= { G m M \over r^2 }$ ? It seems to me that at that speed, you would have to take into account the change o fthe gravitational force with distance.

 

[nrqed]

can anyone else help please? I am positive that i did it wrong and just got lucky

I get 145 km using for gravitational potential enegry $-G m M / r$.

Pat

 

[pureouchies4717]

thank you guys/girls!