PROJECTILE LAUNCHED FROM EARTH'S SURFACE?
A projectile is shot straight up from the earth's surface at a speed of 1.20×10^4km/hr. How high does it go? (answer in km)
Supposed to solve this using F=G(mm/2r)...?
To solve this problem, we need to use the equations of motion for a projectile launched vertically. The equation that relates the height reached by the projectile to its initial velocity is:
h = (v0^2) / (2g)
where:
h is the height reached
v0 is the initial velocity (1.20×10^4 km/hr)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Before we can proceed, we need to convert the initial velocity from km/hr to m/s. To do this, we divide the velocity by 3.6:
v0 = (1.20×10^4 km/hr) / (3.6 km/hr/m/s) = 3.33...×10^3 m/s (rounded to 3 significant figures).
Now that we have the initial velocity in m/s, we can calculate the height reached:
h = [(3.33...×10^3 m/s)^2] / (2 * 9.8 m/s^2) = (1.11...×10^7 m^2/s^2) / (19.6 m^2/s^2) = 5.66...×10^5 m (rounded to 3 significant figures).
Therefore, the projectile reaches a height of approximately 5.66×10^5 meters or 566 kilometers.