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?

Oh, we have a real high flyer here! Well, let me do some calculations while I avoid any birds that might be in the way. So, let's see... if we convert that speed to meters per second (because we don't want to mess with the metric system, right?), we get about 3,330 meters per second. Now, onto the height!

Using some fancy physics, we can determine that the maximum height reached by the projectile can be calculated using the equation h = (v^2)/(2g), where v is the initial velocity and g is the acceleration due to gravity. For simplicity's sake, we'll assume g is approximately 9.8 meters per second squared.

Plugging in the numbers, we get:

h = (3,330^2)/(2*9.8)

And if you'd like a straight answer, well, it goes up approximately 5.85 kilometers. That's pretty high! Just high enough to reach cloud nine, I'd say. But don't forget, what goes up must come down, so keep an eye out for any falling projectiles if you're in the area!

To determine how high the projectile goes, we need to use the equations of motion. However, the given initial velocity of the projectile is in kilometers per hour (km/hr), so we need to convert it to the appropriate units for calculations.

First, we convert the initial velocity from km/hr to meters per second (m/s).

1 km = 1000 m
1 hour = 3600 seconds

So, to convert km/hr to m/s, we multiply the given value by (1000/3600):

1.20 x 10^4 km/hr * (1000 m/3600 s) = (1.20 x 10^4 km/hr * 1000 m) / 3600 s = 3333.33 m/s (approximately)

Now that we have the initial velocity in the correct units, we can proceed to calculate the height reached by the projectile.

The motion of the projectile can be described by the equation:

h = (V₀² - V₂²) / (2g)

Where:
h is the maximum height reached by the projectile,
V₀ is the initial velocity of the projectile,
V₂ is the final velocity at the highest point (which is zero when it reaches the highest point),
and g is the acceleration due to gravity (9.8 m/s² on the Earth's surface).

Plugging in the values into the equation:

h = (3333.33 m/s)² / (2 * 9.8 m/s²)
h = (11111.10 m²/s²) / 19.6 m/s²
h ≈ 566.95 meters

So, the projectile reaches a height of approximately 566.95 meters.

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

That's interesting...because the altitude is so high you have to deal with escape velocity instead of the usual kinematics.

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.

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.