A projectile is shot straight up from the earth's surface at a speed of 1.20 x 104 km/hr. How high does it go?
you can work this from kinemetics, but l like energy...
initial KE= change in PE
1/2 m vi^2=GMe*m (1/(re+1.2E7m) - 1/re)
but GMe/re^2= 9.8m/s^2
so 1/2 vi^2=9.8 re (1/(re+h) - 1/re)
now, change vi to m/s, look up radius of earth re in meters, and solve for h
To determine how high the projectile goes, we need to use the principles of projectile motion and convert the given speed from km/hr to m/s. Here are the steps to follow:
1. Convert the speed from km/hr to m/s:
- Given speed = 1.20 x 10^4 km/hr
- Conversion factor: 1 km/hr = 0.2778 m/s
- Speed in m/s = (1.20 x 10^4) * 0.2778 = 3333.6 m/s (rounded to four significant figures)
2. Use the equation for vertical motion under gravity:
- The vertical motion of the projectile can be divided into two parts: upward motion and downward motion.
- The initial velocity of the upward motion is the speed we just calculated, and the final velocity is 0 m/s since the projectile reaches its maximum height.
- Use the formula: v^2 = u^2 + 2as, where v = final velocity, u = initial velocity, a = acceleration (acceleration due to gravity), and s = displacement.
- Rearranging the equation, we get: s = (v^2 - u^2) / (2a)
3. Determine the acceleration due to gravity:
- The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s^2 (rounded to two significant figures).
4. Calculate the displacement:
- Substitute the values into the formula: s = (0^2 - 3333.6^2) / (2 * 9.8)
- s = (-11,112,889.6) / 19.6
- s = -566,542.7 m
5. The displacement (height) is negative as it represents the distance below the starting point. To find the height above the Earth's surface, we take the absolute value:
- Height = |displacement| = |-566,542.7| = 566,542.7 m
Therefore, the projectile reaches a height of approximately 566.5 km above the Earth's surface.