Follow the steps explained in my previous answer.
http://www.jiskha.com/display.cgi?id=1270449424
If you are just looking for numbers to put into a test or homework, rather than an understanding of how the problem is done, then you have come to the wrong place.
A projectile is shot straight up from the earth's surface at a speed of 1.20×104 KM/HR.
How high does it go?
4 answers
First need to convert km/hr to meters/second
(I got around 3333.33 m/s but you might want to double check that)
so 3333.33 is your initial velocity v_i
using kinematics equation:
v_f=V_i+at
y=v_i*t+(1/2)at^(2)
the highest point the projectile reaches is when v_f is zero
so solve for time using the first equation and the known variables
a= 9.8 m/s^2
v_i= 3333.33 m/s
use the time you determined and plug it into the second kinematics equation and then you'll get your max height.
(I got around 3333.33 m/s but you might want to double check that)
so 3333.33 is your initial velocity v_i
using kinematics equation:
v_f=V_i+at
y=v_i*t+(1/2)at^(2)
the highest point the projectile reaches is when v_f is zero
so solve for time using the first equation and the known variables
a= 9.8 m/s^2
v_i= 3333.33 m/s
use the time you determined and plug it into the second kinematics equation and then you'll get your max height.
The method described above will not give the correct answer because g is not constant over the large altitude range that the projectile will travel.
Refer to http://www.jiskha.com/display.cgi?id=1270449424
Refer to http://www.jiskha.com/display.cgi?id=1270449424
That's interesting...because the altitude is so high you have to deal with escape velocity instead of the usual kinematics.
0 answers