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?
well at the point it is highest all KE=PE, 1/2mv^2=mgh
No you use -GMm/r to do this...
To determine how high the projectile goes, we need to break down the problem into smaller steps.
Step 1: Convert the initial speed from km/hr to m/s.
Given: Initial speed = 1.20×10^4 km/hr
To convert km/hr to m/s, we need to multiply the initial speed by a conversion factor.
1 km = 1000 m (since 1 km is equal to 1000 meters)
1 hour = 3600 seconds (since 1 hour is equal to 60 minutes and 60 seconds)
Conversion Factor = (1000 m) / (3600 s)
So, to convert km/hr to m/s, we multiply the initial speed by the conversion factor:
Initial speed in m/s = (1.20×10^4 km/hr) * (1000 m / 3600 s)
Step 2: Calculate the maximum height reached by the projectile using the formula of projectile motion.
Given: Initial speed in m/s (from Step 1)
The maximum height reached by a projectile launched vertically can be calculated using the formula:
Max height = (Initial speed)^2 / (2 * Acceleration due to gravity)
Acceleration due to gravity on Earth is approximately 9.8 m/s^2.
So, the maximum height reached by the projectile is:
Max height = (Initial speed in m/s)^2 / (2 * 9.8 m/s^2)
Step 3: Substitute the calculated values into the formula and solve for the maximum height.
Now we can substitute the initial speed (from Step 1) into the formula obtained in Step 2:
Max height = [(1.20×10^4 km/hr) * (1000 m / 3600 s)]^2 / (2 * 9.8 m/s^2)
Simplifying this expression will yield the answer for the maximum height reached by the projectile in meters.